bồi dưỡng học sinh giỏi toán tổ hợp rời rạcnguyễn văn thông

'^d gido iCu til - Thqc si Todn hoc - Gido vien chuyen Le Quy Don
N G U Y I N VAN THONG

TOAN
Ti llfP; 111 MC
I

Danh cho hpc sinh chuyen

D M
NAII

I

Toan-„^in^

NHA XUAT BAN DAI HOC QUOC GIA HA NQI


Cty TNHH MTV DWH Khang Viet

mii

Maddu

XURT BAN Dfll HOC QUOC Glfl Hl^ NOI
16 Hang Chuoi - Hai Ba TrUng - Ha Npi
Dien t h o a i : Bien tap - Che ban: (04) 39714896;

Hanh chinh: (04) 39714899: Tona bien tap: (04) 39714897
Fax: (04) 39714899

Tu duy ve toan roi rac ra doi rat som. O Trung Quo'c, vao thoi nha Chu
nguai ta da biet den nhimg hinh vuong than bi.
Nha toan hpc Pithagore va cac hoc tro cua ong da phat hien ra nhieu tinh
chat ky la ciia cac so. M o t ket qua noi tieng cua cac truong phai nay la ke't qua
ma ngay nay chiing ta goi la dinh l i Pithagore. Tuy nhien c6 the noi rang ly

Chiu trdch nhi^m xuat ban
Gidmdoc-

Tong

bien tap :

thuyet toan roi rac duoc hinh thanh nhu mpt nganh toan hoc chi vao khoang
the ky XVII bang mgt loat cong trinh nghien cmi nghiem tiic cua cac nha toan
TS. P H A M T H j T R A M

hoc xuat sac n h u Pascal, Fermat, Leibnitz, Euler...
Chiing ta nho lai hai bai toan noi tieng trong toan roi rac sau:

Bi^n

tap

N G Q C LAM

CM

ban

C O N G TY KHANG V I E T

Trinh

bay

bia

<6di toan 1:

pi
n

B

C O N G TY KHANG VIET

Tong phdt hanh vd doi tdc lien ket xuat ban:

CONG TY TNHH MTV
D|CH Vg VAN H 6 A KHANG VIET

- EULER va bai toan bay cau 6 KONIGSBERG
6 thanh pho KONIGSBERG (LITHUANIA) c6 7 cay cau bac ngang con

/^Dia chl 71 Dinh Tien Hoang - P Da Kao - Q 1 - TP HCM
Dien thoai- 08. 39115694 - 39105797 - 39111969 - 39111968
Fax; 08. 3911 0880
Email: khangvletbookstore®yahoo.com.vn

Website: www.nhasachkhangviet.vn

song pregel n h u hinh ve tren:
Nguai ta dat ra cau do.
^

Tim each di qua tat ca bky cay cau nay, moi cai diing mpt Ian roi quay ve
diem xuat phat.
Nam 1736 Leonahard Euler (1707

SACH L I E N K E T

- 1783) da chung minh khong the c6

BOI DlfONG HQC SINN GIOI TOAN TO HQP - R6I RAC

mpt duong d i nao nhu the bang lap

M a so: 1 L - 2 9 7 D H 2 0 1 2
In 2 . 0 0 0 c u o n , k h d 1 6 x 2 4 c m
T a i : Cty T N H H M T V IN A N MAI T H j N H DLfC
Dja c h l : 7 1 , Kha V g n C a n , P. H i e p Binh C h a n h , Q . Thu Diifc, TP. H o Chi M i n h
So xuat b 5 n : 1 2 0 4 - 2 0 1 2 / C X B / 0 7 - 1 9 4 / D H Q G H N ngay 0 9 / 1 0 / 2 0 1 2 .
Quyet d j n h xuat bkn so: 2 9 2 L K - T N / Q D - N X B D H Q G H N , cap ng^y 09/10/2012
In xong

n o p \\Ju c h i e u quy IV nSm 2 0 1 2

'uan sau:

i^.u, •

Bieu dien bon mien dat A, B, C, D
bang bon diem trong mat phang, moi
cau noi hai mien duoc bieu dien bang
nipt doan noi hai diem tuong ung ta
se

CO

do thi sau:


Cty TNHH MTV

Boi dudng hqc sink gioi Toan tohgrp - red rac, Nguyen Van Thong

Bay gio, m p t d i r a n g d i qua bay cay cau, m o i cai d u n g m p t Ian r o i quay ve
d i e m xua't p h a t se c6 so' Ian d i den A bang so' Ian r o i k h o i A , nghia la p h a i sit
d u n g den m p t so chan cay cau n o i v o i A . V i so cau n o i v o i A la 5 (le) nen

DWH

Khang

Vi^l

p6, to h o p va l y t h u y e t d o t h i lien q u a n chat che v o i n h a u , k h i d o bp m o n Toanj
r d i rac bat d a u p h a t t r i e n m a n h me.
-

' " •

T r o n g t r u o n g p h o t h o n g cua Vi?t N a m , hi?n nay bp m o n nay chua thuc

k h o n g CO d u o n g d i nao thoa m a n d i e u k i e n bai toan. Y t u o n g tren cua Euler da

svf d u p e ehii t r p n g va q u a n tarn, chua c6 h u o n g n g h i e n c u u de d u a toan r o i rac

k h a i sinh n g a n h toan hpc c6 n h i e u ap d u n g d o la l i t h u y e t d o t h i .

vao giang day t r o n g t r u o n g p h o t h o n g cho eo ke't qua. N h a t la d o i v o i cac lop

^di toan2:
Bai toan b o h m a u .
-

>5,.,

...

•f>:v,i
:m -^tn

,;.u:; ...i

^ , ,r
u-^-i.

H '•':•;.;]': i:n^,;piS.jjfin^f> ;:;firr, y j l ^ i i

^

Tren m p t ban d o bat k i , ta n o i no d u p e to m a u , neu m o i m i e n ciia n o

d u p e to m p t m a u xac d i n h sao cho hai m i e n ke n h a u c6 m a u sac khac n h a u .
Van de can d u n g t o i t h i e u bao n h i e u m a u de to m p t ban d o bat k i .
Bai toan da d u p e dat ra k h o a n g g i u a the k i XIX.
M o h i n h toan hpc cua bai toan nay n h u sau:
V o i m o i ban d o M cho trude, ta hay xay d u n g m p t d o t h i G t u o n g u n g , m o i
m i e n cua M u n g v o i m p t d i n h cua G, 2 m i e n ciia M c6 c h u n g m o t p h a n bien
neu va chi neu 2 d i n h t u o n g u n g t r o n g G c6 canh n o i . De thay do t h i G n h a n
d u p e theo each tren la m p t d o t h i phang n h u the bai toan to m a u M t r o t h a n h
bai toan to m a u G . N g a y tir k h i m o i xua't hien bai toan n g u o i ta dat gia thie't bai

chuyen Toan ehuyen T i n , n h i m g l o p p h o t h o n g chuyen can b o i d u o n g Toan r o i
rac de cac e m vao d a i hpc, hpc bp m o n nay tot h o n .
-

N g a y t r o n g cac k y t h i HSG quoe te^ hpc sinh Vi?t N a m ehiing ta t h u o n g

khong thanh cong v o i cac bai toan To h p p , to m a u , do t h j . T r o n g t i n h h i n h toan
hpc ngay cang phat trien, t h i su ke't h p p g i i i a cac ITnh v u c toan hpc v o i nhau
ngay cang cao va sau sac, d o i h o i n g u o i l a m toan t u o n g lai p h a i biet n o i ke't v o i
cac bp m o n lai v o i nhau m o i dat ke't qua m o n g m u o n , m a bp m o n toan r o i rac
pho t h o n g la buoe d a u can thie't. V o i t u d u y sau sac, m a n h me, n o se la vien gach
dau tien cho hpc sinh p h o thong ren l u y e n buoe d a u d i vao nghanh toan hpc va
mpt t r o n g so hp se t r o thanh nha toan hpc ehuyen n g h i f p t u o n g lai.
-

D a y la ke't qua t h i O l y m p i c toan quoe t e c u a doan hpc s i n h V i ? t N a m Ian

t h i i 46 tai M e x i c o n a m hpc 2005 (trich t u toan hpc va t u o i tre so 8/2005).

toan la 4. T u y n h i e n , k h o n g eo m p t c h u n g m i n h nao d i i n g cho gia thie't nay
d u p e dat ra. C h o den n a m 1976

m p t n h o m cac nha khoa hpc ( K . A p p e l ,

SBD^\

BAIl

BAI 2

BAI 3

BAI 4

BAI 5

BAI 6

W . H a k e n , J.Kock) da xay d t m g m p t l o i giai d u a tren cac ke't qua d o m a y t i n h

VNl

5

7

0

7

7

4

I B M 360 c u n g cap (mat h a n g n g a n g i o tren m a y t i n h ) da k h a n g d i n h gia t h u y e t

VN2

0

1

7

2

7

2

VN3

0

7

0

1

7

0

VN4

1

1

6

1

7

0

VN5

7

7

6

2

7

2

VN6

7

7

7

2

7

2

E

20

30

26

15

42

10

4 m a u la d u n g .
-

C l i n g v o i s u phat trien, v o i toe d p n h a n h ciia cong nghe t h o n g t i n , l y

t h u y e t to h p p v a d o t h i da t r a t h a n h m p t l i n h v y c toan hpc q u a n t r p n g va can
thie't cho n h i e u l i n h v u c khoa hpc va u n g d u n g .
-

Cac hpc sinh, ke ca thay giao v a n eon chua q u a n tarn n h i e u den p h a n

nay: l y t h u y e t g i , bai tap nao la can thie't cho viee b o i d u o n g H S G Toan, HSG
Tin, HSG t r o n g t r u o n g T H P T .
-

C h u a CO cong t r i n h nao n g h i e n c u u that sau sac va day d i i cho hpc sinh

p h o t h o n g ve cac u n g d u n g cua toan hpc r o i rac.
-

Mac d a u ve m a t lich su, toan r o i rac ra d o i tu hai t r a m n a m t r u d c day

t r o n g qua t r i n h giai cac bai toan do, n h u n g bai toan d a n gian. T u lau, n o v a n
con d u n g n g o a i le cac p h u o n g h u o n g n g h i e n euu cua cac nha Toan hpc, den

nay v i t r i cua no d u p e k h a n g d i n h t r o n g Toan hpc, dac biet la t i n hpc h i e n d a i .
m^nh

Buoe p h a t t r i e n cua toan r o i rac, da t h u d u p e n h i i n g ke't qua va phat t r i e n
vao k h o a n g euoi t h e k i 19 va d a u t h e ' k i 20, k h i m a cac cong t r i n h ve to

(Bangl)

^

Bai 6 la bai toan r o i rac t r o n g k y t h i I M O k y 46 tai M e x i c o c6 n p i d u n g sau:
T r o n g m p t k y t h i hpc sinh gioi, cae t h i sinh p h a i giai 6 b a i toan. Biet rang hai
2
%. bai toan bat k y l u o n c6 n h i e u h o n - so t h i sinh d u t h i giai d u p e ca hai bai nay.
o

N g o a i ra k h o n g c6 t h i sinh nao giai d u p e ca 6 bai toan. C h u n g m i n h rang c6 i t
nhat 2 t h i sinh sao cho m o i n g u o i t r o n g hp giai d u p e d u n g 5 bai toan.
- Bai nay ciia R u m a n i de n g h i . (Doan hpc sinh V i ? t N a m dat ke't qua thap).


Boi duanghQc sinh gioi Toan tohgp-rari

r^c, Nguyeu

//••,;

Cty TNHH MTV DWH

Day la ket qua ky thi Olympic toan quo'c te cua doan hpc sinh Vi^t Nam Ian

thii 47 tai Slovenia nam hpc 2006 (trich t u toan hpc va tuoi tre so'8/2006).

Khang Vi?t

. Qua qua trinh giang day cac lop chuyen Toan + Tin va ket hpp voi
chuong trinh boi duong HSG cua Bp giao due va dao tao toi viet tai lieu nay
vdi nhung kinh nghif m ban than ma toi da day cong nghien euu trong suot
qua trinh giang day. Trong tai lieu nay, toi eiSng manh dan dua ra mot so khao
sat mdi cua minh de tong quat hoa mpt so ket qua da biet.

Bail

Bai 2

Bai 3

Bai 4

Bai 5

Bai 6

VNl

7

0

1

7

3

0

VN2

7

1

7

6

7

0

T6NGQUAN

VN3

7

1

7

7

7

0

Quyeh sach nay se dupe chia lam sau chuong.

VN4

7

0

1

6

1

0

Chuong 1: A n h xa va luc lupng ciia cac tap hpp.

VN5

7

0

7

7

1

0

Chuong 2: Phep dem, to hpp, chinh hpp va cac mo rpng.

VN6

7

4

1

7

0

0

Chuong 3: Trinh bay ve do thi, cac dinh nghia dinh ly can ban thuong dung

Z

42

6

24

40

19

0

S B D \

(Bang 2)
Bai 2 va Bai 6 la hai bai toan roi rac do nude (Serbia) de nghi. Ket qua ciia
doan Viet Nam rat thap trong ky I M O Ian thu 47 to chuc tai Slovenia 2006.
Den nay ket qua I M O ciia chiing ta van chua c6 dau hieu tien bo, ma hpc
sinh van con lam tot cac bai toan Pi va Ps chuyen ve hinh hoc phang, con bai
toan 3, bai toan 6 van con rat yeu, bai toan 3 ve to hop roi rac, bai toan 6 thupc
ve so hpc. Chiing ta c6 the quan sat ket qua doan Viet Nam dual day.
STT

Hp va ten

PI

P2

P3

P4

P5

P6

:f^K'rWl':,:~::^ly/,r^,.. r i ,
, X i ;j ; i

cho hpc sinh chuyen Toan theo gioi han chuong trinh cua Bp giao due.
Chuong 4: Trinh bay ve to mau gom nhung khai niem eo ban, cac dinh ly
thuong su dung trong giai Toan pho thong theo gioi han chuong trinh ciia Bp
giao due.
Chuong 5: Toan to hpp va thi HSG

y g / \

Chuong 6; Nguyen ly Dirichlet - Djnh ly Ramsey

^ ,,

,

- Trinh bay theo quan diem ly thuyet voi nhung chung minh dinh ly, vi du
minh hpa, bai tap ap dung, bai tap t u luyen.
- Bai tap chpn Ipe t u miic dp trung binh eho den kha, gioi, cac de thi toan

Tong

Huy

diem

chuong

1

Nguyen Ta Duy

7

7

0

6

7

0

17

Bac

2

Dau Hai Dang

7

7

3

7

7

0

31

Vang

3

Nguyen Phuong Minh

7

7

0

6

7

0

27

Bac

4

Le Quang Lam

7

7

0

5

0

0

19

Dong

5

Tran Hoang Bao Linh

7

1

0

5

7

0

20

Dong

6

Nguyen Hung Tarn

7

7

1

2

7

0

24

Bac

(Bang 3 (Trich Toan hpc va tuoi tre so 8 - 2012))
- Qua nhiing ky thi dinh cao, nhu vay, thi sinh d u thi toan la nhung hpc

quoe gia, quo'c te.
- Chiing minh nhieu bai toan chia khoa, de nhin nhan van de don gian hon.
- Cac dinh ly ve to hpp thuong su dung phuong phap anh xa, phuong
phap song anh de chiing minh.
- Cac bai toan ve do thi dua vao logic de suy luan va chiing minh eo gang
dua ra nhieu bai toan chia khoa de hpc sinh de ap dung va van dung.
Day la van de kho nhung tac gia mong muon gop phan minh vao ky nang
lam toan roi r^c cho hpc sinh chuyen toan nen khong khoi tranh sai sot. Rat
mong gop y ciia cac ban dong nghiep va cac em hpc sinh.

.. ,j .,, ^

sinh gioi dupe chpn Ipc ky luong nhung nhin vao ket qua thi ta nhan tha'y.
- Hpc sinh Vi?t Nam rat manh ve hinh hpc p h i n g , bat dang thuc, giai tich,
phuong trinh ham, nhung toan roi rac van la diem yeu. Do do van de dat ra la

Nguyen Van Thong

phai CO ly thuyet va bai tap ve toan roi rac de boi duong cho hpc sinh gioi toan
pho thong, hpc sinh cac lop chuyen toan, chuyen tin, can boi duong can than
va day dii de hpc sinh nam dupe ban chat van de va kha nang ling dung, theo
duoi dupe con duong toan hpc sau nay.
6

7


BSi duang hgc sink gidi Toan

td'ht^

- rbi rac, Nguyen

Van

Thdng

^inh

nghla 1.3.3.

Ta bao m p t anh xa f: X

A N H XA, L U CLUCWG C U A C A C T A PH O T

X len Y, ne'u no vira la d o n anh vira la toan anh, n o i m p t each khac ne'u v o i m p i
y eY

§1. A N H X A

1.1.1.

•

'

>'

^inh

'
s

»

cr-"

f: X

Anhx?

Y v a g: Y -> Z.

X->Z

'«

>

•

'

'^'f •

x-^g(f(x))

1.2. A N H V A T A O A N H :
^inh

nghla 1.4.1.

Gia s u cho

M p t anh xa f t u X vao Y la m p t q u y tac cho t i r o n g u n g v a i m o i p h a n t u x e X
m p t p h a n tvr xac d i n h y e Y, k y hi?u f(x).

m p t v a chi m p t x e X sao cho y = f (x)

1.4. TfcH A N H X A .

'
•'

CO

Chang han anh xa d o n g nhat I x la m p t song anh v o i m p i X.

I.I.OINHNGHTAANHXA:
^inhnghia

Y la m p t song anh hay m p t anh xa m p t do'i m p t t u

.^
I'M

_ » G p i la tich anh xa f v a anh xa g. K i hieu: g.f

nghla 1.2.1.

^inhly

Gia s u f: X -> Y la m o t anh xa da cho, x la m o t phan t u t i i y y cua X, A la mQt
bp phan t i i y y ciia X, B la m p t b p p h a n t i i y y ciia Y. The t h i n g u o i ta gpi:

' 1 "

1.4.2.

Gia s u cho
The t h i

f: X ^ Y, g: Y, g: Y ^ Z, h: Z ^ T

3'- W '

h(gf) = (hg)f.

^ y'.Ji^iich •

•

f(x) la anh cua x b o i f hay gia t n ciia anh xa f tai d i e m x.

Ta bao phep n h a n cac anh xa c6 t i n h chat ke't h o p .

•

f ( A ) = { y e Y I t o n tai x e A sao cho f(x) = y ) la anh ciia A b o i f.

C h u n g m i n h : Ta CO v o i m p i x e X.

•

f"^ (B) = { X e X I f(x) 6 B 1 la tao anh toan p h a n cua B b o i f.

(h(gf)) (x) = h(gf(x)) = h (gf(x)) = (hg)((f(x)) = ((hg)f){x).

Dac biet v o i b e Y, f"^ ({ b }) = { x e X I f(x) = b }. De d o n gian k y h i e u ta viet
(b) thay cho f"^ ({ b}) v a g p i la tao anh toan p h a n ciia b b o i f. M o i p h a n t u

K y . h i f u £(A) la m o t d i e u l a m d u n g v i f(A) chi c6 nghla k h i A e X. R6 rang ta
CO f ( 0 ) = 0 v o i m o i f. Ta c h u n g m i n h de dang cac quan he:
•
v»

•

-

s:.v, ^.T

,inur-

gf = l x v a f g = l y

^inh

nghla 1.4.4.

A n h xa f: X

nghla 1.3.1.

A n h xa f: X -> Y la m o t d o n anh ne'u v o i m p i x, x ' e X, quan h ^ f(x) = f(x')
f(x'), hay v o i m p i y e Y c6

nhieu nhat m p t x e X sao cho y = f(x). N g u o i ta con g p i m p t d o n anh f: X

Y

la m p t anh xa m p t d o i mot.
nghla 1.3.2.

Ta bao m p t anh xa f: X ^ Y la m p t toan anh ne'u f(X) = Y, n o i m p t each khac,
ne'u v o i m p i y e Y c6 i t nhat m p t x e X sao cho y = f(x). N g u a i ta con g p i m p t
toan anh f: X ^ Y la m p t anh xa t u X len Y.

'

D i n h l y sau day cho ta biet k h i nao m p t anh xa c6 anh xa ngirpc.

1.3. D O N A N H - TOAN A N H - S O N G ANH:

^inh

1.4.8.

Gia s i i f: X - » Y v a g: Y - » X la hai anh xa sao cho

Tir d i n h nghia ta suy ra f cung la m p t anh xa ngupc ciia g.

B c f ( r ^ (B)) v o i m p i b p phan B ciia Y.

keo theo q u a n h$ x = x', hay x ^ x ' keo theo f(x)

C h i i y rang neu f: X -> Y la m p t anh xa bat k y t h i ta c6: fix = l y f = f

The t h i g g p i la m p t anh xa ngupc ciia f .

A c f"-^ (f(A)) v o i m p i b p phan A ciia X.

^inh

•

D o d o ta k y h i e u h(gf) = (hg)f bang h g f va g p i la tich ciia ba anh xa f, g, h .
'i>inhnghla

x e f"^ (b) g p i la m o t tao anh ciia b b o i f.

Y CO m p t anh xa ngupc k h i v a chi k h i f la m p t song anh.

Chung mink G i a sii- f c6 m p t anh xa ngupc g: Y -> X
Theo d i n h nghia ta c6:

gf = I x v a fg = ly

Tucla:

g(f(x)) = x v o i m p i x.

-

Xetquanhe:

f(x) = f(x')

Tasuyra

x = g(f(x)) = g(f(x')) = x'.

,
v , ,

Vay f la m p t d o n anh. Bay g i o gia s u y la m p t p h a n t u t i i y y ciia Y. D a t x =
g(y) e X t r o n g d a n g t h i i c f(g(y)) = y, ta dupe y = f(x). Vay f la m p t toan anh.
Dao lai, gia s u f la m p t song anh. Q u y tac cho h r o n g u n g v o i m o i y e Y phan
t i i d u y nhat ciia f"^ (y) xac d i n h m p t anh xa g: Y

8

.f

X va ta thay ngay
9


Boi duong hQc sink gidi Todn

tohipp

- rai rac, Nguyen

gf =

I x

Van Thdnjj

Cty TNHH

va fg = ly

g(y) =

X,

sao cho f(x) = y

g

1. H a i tap

Y.

T h e t h i g = g'.

^

T u do:

gf = I x va fg'= ly

'

'

i

s

it; =

, '

(y) bang

i ; ; ^

(f''^)"^.

Vay

/

I

I

\

/

v a i tap h i i u han).

V 1

trong m g t khoang (a, b) bat ky: s u
t u a n g l i n g c6 the thuc hien bang

Jib

Hinh 1

hay cung m g t ban so'. N o i each khac luc l u g n g (hay ban so) ciia m g t tap bieu
t h i m o t t i n h chat chung cho no va tat ca cac tap t u o n g d u a n g v a i no. D o i v a i

= Ix.

tap hijoi han, t h i luc l u g n g (ban so) chinh la so' phan t i i cua no. Thanh t h i i , luc
l u g n g (ban so) la k h a i n i e m true tiep m o rgng khai n i e m so' t u nhien. Luc
l u g n g ciia m g t tap A t h u a n g d u g c k i hieu I A I (hay cardA).

C h u n g m i n h . Ta c6:

(g-'

/

K h i hai tap h g p t u o n g d u o n g nhau, ta bao rang c h i i n g c i i n g m g t luc l u g n g
nen f =

Ji^ qua Cho hai song anh f: X ^ Y v a g: Y - » Z. The t h i gf: X -> Z la m g t
song anh.

(gO(

I I \
I \

•

phep v i t u ( H i n h 1)

y ^ f - l (y)

CO

tu

(0,1) t u a n g d u o n g v a i tap cac d i e m

(y) va d o d o n g u a i ta k y h i e u anh xa

Ta

S

3. T a p cac d i e m trong khoang

D o l a m d u n g n g u a i ta cung k y hieu phan t u d u y nha't x ciia

V i f la anh xa n g u g c ciia

t u o n g d u o n g v i c6

't

ciia n o (dieu k h o n g the xay ra d o i

«.

f"^.

2n,

(n = l , 2 , 3,...)

d u a n g v a i m g t b g phan thuc s u

y ^ X , v o l X la p h a n ttr d u y nhalt ciia f"^ (y)

la anh xa n g u g c ciia f, bang

n, ...} va B = { 2, 4, 6,

vay m g t tap v 6 han c6 the t u a n g

'

N h u vay neu f: X ^ Y c6 anh xa ngugc t h i anh xa n g u g c la d u y nha't, xac
dinhboi.

h a n c i i n g so'lugng t h i t u o n g d u o n g .

D i e u dang chii y 6 day la B c A:

' g = gly = g(fg') = (gOg' = l^g' = g'-

^

Vi^t

the thiet lap phep t u o n g d u o n g l i n g :
n<^2n

<^inh ly 1.4.5. Gia su g: Y ^ X va g': Y -> X la hai anh xa ngugc ciia f: X
c6:

hiJTi

2. H a i tap A = { 1, 2, 3,

N g o a i anh xa ngugc nay, f con c6 anh xa ngugc nao khac khong? Ta c6:

ChicngminhTa

Khang

Vidu:

N h u vay f: X ^ Y c6 m p t anh xa ngugc k h i v a chi k h i f la song anh, va
t r o n g t r u a n g h g p do ta c6 m p t anh xa ngugc g: Y
X ciia f xac d i n h b o i
>i ,
Y

MTV DWH

i

g-^) = g ( f f - ^ ) g - i = g i y g - i = g g - i = i .

)(gf) =rU g"' g)f = r' hi =r'i=ix

>

;

f-ig-^=(gf)-i

N e u m g t tap A t u o n g d u o n g v o i m g t bg phan ciia m g t tap B, n h u n g k h o n g

t u o n g d u a n g v a i B, t h i ta noi luc l u g n g ciia A nho h a n luc l u g n g ciia B, hay
lye l u g n g ciia B Ion h a n luc l u g n g ciia A va vie't I A I < IBI hay IBI > I A I .
M g t v a n de t u nhien nay ra: neu m g t tap A t u o n g d u o n g v a i m g t bo phan
ciia tap B va n g u g c lai tap B c i i n g t u o n g d u a n g v a i m o t b g phan ciia A , t h i eo

§2.

L l J C Ll/ONG CUA CAC TAP HOP

the n o i gi ve h a i tap ay? D i n h ly sau day giai dap v a n de nay.
2.1.1. ^inh ly (Cantor- Berntein). N e u tap A t u a n g d u o n g v a i m g t bg phan

2.1. TAP HOP TUONG O JONG:
K h i ta d e m Ian l u g t cac phan t u ciia mgt tap A, c6 the xay ra hai t r u a n g hgp:
1. T a i m g t liic nao do, ta d e m d u g c het cac p h a n t i i ciia A . T r o n g t r u o n g
h g p nay, tap A la h i i u han v a so cuoi c i i n g d a d e m t a i cho ta biet so l u g n g
phantuciiaA.
ly,
•'
.>vi
2. M a i m a i v a n con n h i i n g phan t i i ciia A chua d e m t o i . T r o n g t r u a n g h g p
nay, tap A la v 6 han.
Ta n o i hai tap h g p A v a B la t u a n g d u o n g (ve so' l u g n g ) neu g i i i a hai tap
h g p ay c6 the thiet lap m g t phep t u a n g l i n g 1 - 1 (tiic la c6 the anh xa 1 - 1 tap
nay len tap kia).

'

ciia tap B v a n g u g c lai tap B c i i n g t u o n g d u o n g v a i m g t bg p h a n ciia tap A t h i
hai tap t u a n g d u a n g nhau.

N h u vay, neu tap A t u o n g d u o n g v o i m g t
bg phan ciia tap B t h i chi c6 the I A I < IBI hoac
l A I = IBI (ta vie't l A I < I B I ) . N o i each khac.
l A I < I B I , IBI < l A I
Chung

mink

l A I = IBI

theo gia thiet c6 m g t phep t u o n g

li'ng 1 - 1 f giiia cac p h a n t u ciia A v a i cac phan
t u ciia m g t bg p h a n ciia B, va ngugc lai cung c6

Hinh 2
11


BSi duang htfc sink gioi Todn tohgrp - rcri r^c, Nguyen

Vin

C t y TNHH

Thdng

M T V DWH

Khang

Vift

mpt p h e p tirong ung 1- 1 g giiia cac p h a n t u
cua B vai cac phan t u ciia mpt bp phan ctia A . s'< <
^ii ; ' ;
;
Ta qui uoc gpi mpt phan t u x la "ho" cua phan t u y hay y la "con" cua x,
neu x e A , y G B v a y = f(x) (y l i n g voi x trong phep tuong ung f) hoac neu xe B,
y e A va y = g(x) (y ung vdi x trong phep tuong ung g).

Vi du;Theo v i d\ 1 (a myc truoc), t a p |2, 4, 6,
2n, ...} la dem dupe. Ta
cung thay ngay rang cac t a p { 3, 6, 9 , 3 n , ...), { 1 , 4, 9 , n ^ , . . . } , v.v ... deu dem
dupe; t a p hpp cac so nguyen (bao gom ca so nguyen duong, am, va so khong)
C l i n g la dem dupe v i c6 the viet dual dang:
4;,; ,
0,1, -1, 2, -2, 3, - 3 , n , - n , . . .

Mpt phan t u x (thupc A hay B) gpi la "to tien" ciia mpt phan t u y, neu c6 mpt
day XI, X2,
XK sao cho x la bo cua xi, xi la bo cua X2, X2 la bo ciia X3,
XK la bo
ciia y. Chang han trong hinh 2, x la to tien cua y dong thai x ciing la to tien ciia
XI, X2, X3; XI la to tien ciia X2, xs; y, v.v... Bay gio ta hay chia A ra lam ba tap con:
Ac gom tat ca cac phan t u ciia A c6 mpt sochSn to tien, A i gom tat ca cac phan
tu ciia A CO mpt so le to tien va A~. gom tat ca cac phan t u ciia A c6 v6 so' to tien.
Dong thoi ta cung theo each do chia B ra lam ba tap: Be, Bi, B~. Sau do ta xac
dinh mpt phep tuong ung cp nhu sau giua cac phan t u ciia A voi eae phan t u ciia
B: voi moi phan t u thupc Ac hay A - thi cho ung con eiia no, vai moi phan t u

thupc A i t h i cho ung bo ciia no.
(i
r? spi;:;

Viec luc lupng dem dupe la lue lupng be nha't eiia cac t a p v6 han dupe xac

De thay rang (p la mpt phep tuong ung 1 - 1 giua A va B. That vay, trong
phep tuong ung do, voi moi x e A d l nhien ung mpt phan t u duy nhat y e B.
Ngupc lai lay mpt phan t u x nao do thupc A i ; neu y e Bi thi bo ciia no thupc
Ac, nghia la y la eon ciia mpt phan A i ; neu y e Bi thi bo ciia no thupc Ac, nghia
la y la con ciia mpt phan t u x nao do thupc Ac; con neu y e B - thi bo' ciia no
thupc Aoo, nghia la y la con ciia mpt phan t u x nao do thupc A - . Thanh t h u moi
phan t u y e B ung voi mpt phan t u duy nhat x e A trong phep tuong ling (p.
Vay (p la mpt phep tuong l i n g 1-1 giua A va B.
Tir dinh ly nay ta suy ra rang cho truoc hai tap A , B, chi c6 the xay ra mpt
trong 4 truong hop:

•j;

1.

lAI

= IBI

2.

l A I < IBI

3.

IBI < l A I

•

K''^'

->niVx F]

Chung minh; That vay, cho M la mpt tap v6 h^n. Ta hay lay mpt phan t u bat
ky, ai 6 M , roi mpt phan t u a2 e M \, roi mpt phan t u as e M \, 32), v.v...
Vi M v6 h?n, nen dieu do c6 the tie'p tyc mai va ta thu dupe tap dem dupe (ai,
a2, as,...} c

M.

^inh ly 2.2.2. Mpt bp phan eiia mpt t a p dem dupe thi phai la h i m han hay
dem dupe.
Chung mink That vay, cho B la mpt bp phan ciia tap dem dupe A = { ai, a2,
as,...}. Gpi
la phan t u dau tien ciia B ma ta g a p trong day { 8 1 , 8 2 , 8 3 , . . . } , a^^
la phan t u t h u hai ciia B trong day so'do, v.v... Neu trong eae so' 11,12, ••• c6 mpt
so Ion nhat, v i du ip, thi B = {

, a^^a,^ ] la hihi han. Con neu trai lai day

, a i 2 k e o dai v6 tan thi B = (

,a^^.....} la dem dupe.

.XA& •

4. A khong tuong duong voi B, hoac bo phan nao ciia B, va B cung khong

la t a p dem dupe.

^* "

Chung mink Cho mpt day tap dehi dupe: A i , A2, A3... Ta hay viet eae phan tu
eiia moi tap ay thanh mpt day va xep dat cac day thanh mpt bang nhu duoi day.
R6 rang tren moi duong cheo (c6 miii ten) chi eo mpt so hiiu han phan tu: tren
chung tren duong cheo thii n chi c6 n phan tu (eae apq ma p + q = n +1).
Ai

A3

2.2. TAP OfM OLTOC
Trong tat ea cac tap v6 han thi tap "be nhat" (c6 lue lupng kem nhat) la tap
cac so t u nhien: N * = { 1, 2, 3,
n,...}. Lue lupng eiia tap nay gpi la lue lupng
dem dupe, va mpi tap tuong duong voi no gpi la tap dem dupe. Duong nhien,
cung CO the noi: Mpt tap dem dupe la mpt tap ma ta eo the danh so dupe cac
phan t u ciia no thanh mpt day v6 han: a i , a 2 , a 3 , a n , . . .

^

^11

^

^12

>^^13

^332

^

^^

^14

^23

324

^33

334

Veiy ta CO the Ian lupt danh so cac phan tu" tren duong eheo thii nhat, roi
den duong cheo t h u hai, v.v...
^inh ly 2.2.4. Khi them mpt tap hpp hixu han hay dem dupe vao mot tap v6
hcin

'

'

^inh ly 2.2.3. H p p ciia mpt hp huu han hay dem dupe t a p dem dupe cung

tuong duong voi bp phan nao ciia A .

12

'

duong cheo thu nhat chi c6 a n , tren duong eheo thu hai chi c6: 821,812, v.v... noi

^ J"r?>v
^iyjy^ .(;•

dinh boi hai dinh l i sau day:
^inh ly 22.1. ^ai cu t a p v6 han nao cung c6 mpt bp phan la t a p dem dupe.

M , ta khong lam thay doi luc lupng ciia t^p M .
13


Cty TNHH MTV D W H Khang Viet

Boi duang hoc sink gidi Todn tohgrp - rcri r^c, Nguyen Van Thong

Chung minh:Cho N la tap thu dugc khi them vao M mQt tap A hiru han hay
dem dxxoc. Theo Dinh ly 2.2.1 c6 the lay mot tap dem dugc B cz M . Dat M ' =
M\B, ta CO M = M ' u B, N = M ' u B u A. Theo dinh ly truac B u A ciang la dem
dugc, cho nen c6 the lap mot phep tuong ung 1-1 giiia B va B u A. Sau do chi
can lap mot phep tuong ung 1- 1 giiia M ' va chinh M ' , ta se c6 mot phep tuong
ling 1-1 giiia M va N . N h u vay M va N cung luc lugng.
Theo dinh ly nay, ta thay rang tap cac diem thugc mot khoang (a, b)

tuong duong voi tap cac diem thugc doan [a, b]. Vay tap cac diem thugc doan
[a, b] tuong duong voi tap cac diem tren toan duong thang.
^inh ly 2.2.5. Tap tat ca cac day hiiu han c6 the thanh lap dugc voi cac phan
t u ciia mot tap dem dugc la dem dugc.
Noi ro hon, cho A =
day

CO

dang (a^^,

{31,82,33,...}

HJ^ , . . . , 3 ; ^

la mot tap dem dugc, S la tap tat ca cac

) trong do m la mgt so' t u nhien bat ky, a^^ (k = 1, 2,

m) la nhiing phan t u (khong nhat thiet phan biet) ciia A. Ta khang dinh
rang S la dem dugc.
Chung mink Ggi Sm la tap cac day gom dung m phan t u ciia A.

2.3. Ll/C LUQNG CONTINUM:
Qua cac v i du tren kia, ta da di t u tap cac so' t u nhien den tap cac so' h i m ti,
roi den tap cac so dai s6^ moi tap bao gom v6 so phan t u moi so voi cac tap da
xet truoc, the ma luc nao ta ciing chi c6 nhiing tap de'm dugc. Vay thi c6 tap
nao khong de'm dugc khong?
^inh ly 2.3.1. Tap cac so thuc la khong de'm dugc.
Chung mink V i tap cac diem thugc doan [0, 1] tuong duong voi tap cac

diem tren toan duong thSng, ta chi can chung minh rang tap cac diem thugc
do^n [0,1] la khong dem dugc.
Gia sir trai l^i rang tap do de'm dugc, nghla la c6 the danh so' thanh day:
Xi,x2,X3,...

ta hay chia doan [0, 1] thanh ba doan bang nhau. Trong ba doan do

phai CO mgt doan khong chua x^: cho doan ay la

. Ta lai chia

ra ba doan

bang nhau. Trong ba doan do phai c6 mgt doan khong chua X 2 : cho doan ay la
A2 . Ta lai chia A 2 ra ba doan bang nhau, v.v. Tie'p tuc mai, ta se c6 mgt day doan
Aj

A2

3>

A3

3

Nen theo Dinh ly 2.2.3 chi can chung minh rang moi Sm la de'm dugc. Voi

va voi x^ g A ^ . V i I A „ I - > 0 (n -> °°) nen do la mgt
3
day doan that lai va theo nguyen ly Cantor, phai c6 mgt diem ^ chung cho tat

dieu do hien nhien v i rang

ca cac dogn ay. Co' nhien ^ e [0, 1]. Vay ^ phai triing voi mgt x^^ nao do.

'

ViS=U-.iS„

(hie

= A. Ta hay gia thiet dieu do diing voi

de'm dugc), va chung minh cho S^^^.

(Hj^ ,3(2 , . . . , 3 ; ^ , 3 k ) .

Giiia s[^+i va

ro rang c6 su tuong ung 1-1.

(aii'ai2'aim'3k)<^(aii.ai2'-.ai^)

y:

I A„

1=

Nhung ^ 6 A „ voi mgi n, cho nen x^^ e A^^ . Dieu nay trai voi each xay dung

Cho ak la mgt phan t u xac dinh ciia A, va s|^+i la tap tat ca cac day so c6
dang

Voi do dai

3

Ma Sn, da de'm dugc thi SJ^+i cung de'm dugc. Do do S^+i ciing de'm dugc,

cac doan A „ . Do do gia thiet rSng tap cac diem thugc [0,1] de'm dugc la v6 ly.
Dinh ly tren cho thay rang luc lugng ciia tap so thuc Ion hon luc lugng de'm
dugc. Nguoi ta ggi no la luc lugng continum hay luc lugng c.
©mh ly 2.3.2. (Cantor). Cho bat cu tap A nao thi tap tat ca cac bg phan cua
A ciing CO luc lugng Ion hon luc lugng cua A.
Chung mink Ggi m la tap tat ca cac bg phan cua A. Chang han ne'u A= {a, b,

Ji$ qua l : T a p tat ca cac da thuc P(x) = ag + 3 j X

+ ... +

3 ^ x " (n bat ky) vol cac

h^ so ao,3i,...,a„ huu ti, la de'm dugc.
That v^y, cac da thuc do hrong ung 1-1 voi cac day so hihi t i (ao,ai,...,an).
Vi tap cac so'him t i la de'm dugc nen theo dinh ly tren, tap cac da thuc do ciing
de'm dugc.
J f | qua 2:Tap cac so dai so la de'm dugc.
Mgt so dai so la nghi^m so'ciia mgt da thuc voi r\hung he so'hiiu ti. V i tap
cac da thuc nay la de'm dugc, ma moi da thuc chi c6 mgt so' h i i u han nghi^m

s6^ nen tap cac so d^i so' chi c6 the la hiru han hay de'm dugc. Nhung no khong
the h i m h^n v i no bao ham tap cac so'tu nhien, vay no phai de'm dugc.
14

c} thi M gom c6: 0 , {a}, {b}, {c}, (a, b}, {b, c}, {a, c}, A.
Truoc he't ta hay chung minh rang A khong tuong duang voi M . Muo'n the,
gia sir trai lai rang A tuong duong voi M , va cho f la phep tuong ung 1-1 giiia
hai tap ay. Voi moi phan tir x e A ung mgt phan tir xac dinh cua M ma ta se ky
hieu la f ( x ) ; ngugc lai, moi phan tir aia M la f(x) cho mgt phan tir xac dinh x e
A. V i f(x) la mgt phan tir cua M , tuc la mgt bg phan cua A nen c6 the xay ra
tinh trang x e f(x) . Ta se ggi mgt phan tir x sao cho x e f(x) la "xa'u so". Con
ne'u x i f(x) thi x ggi la "tot so". Ggi S la tap tat ca cac phan tir tot so cua A. Vi
rang S e M nen S = £(xo) voi mgt XQ e A. Ta thir xem XQ la xa'u so hay tot so?.
Ne'u XQ la xa'u so thi XQ e f( XQ ) = S, v6 ly v i S chi chira nhimg phan tir tot so.
15


Boi dumtg hpc sink gidi Todn td'hgp - red n^c, Nguyen Van Thong

C o n neu X Q la to't so' t h i XQ e f( X Q ) = S : c u n g v 6 l y v i X g da to't so' t h l p h a i thupc

quen v a i B, d o d o A; va B c6 d i i n g 2 n g u d i quen chung, m p t t r o n g h a i n g u o i
do la A '

S. M a u t h u a n d o c h u n g to r a n g A k h o n g the t u a n g d u o n g v d i M .

n g u d i k i a la m p t Bj nao d o la m p t t r o n g n h u n g n g u d i quen B.

phan ctia M . That vay,

Mhvr v a y m o i Aj d a t d u p e t u o n g u n g v d i m o t B, nao d o . N e u i # j n g h i a la A j ,

gpi M ' la tap cac bp phan ciia A chi g o m m p t p h a n t u d u y nha't t h i M ' c M va ro

A khae n h a u t h i Bj, Bj la n g u d i quen ciia h p cung khae n h a u . ( K h o n g n h u vay

rang M ' va A t u o n g d u o n g nhau, v i m o i x e A c6 the cho i m g v d i tap {x} e M ' .

thi A va Bj, ed ba n g u d i quen c h u n g la B, A j , A j ) . D o t i n h d d i x i i n g , ro rang

Bay g i o ta c h i m g m i n h rang A t u o n g d u o n g v o i m g t

if

T o m l a i , I M I > I A I , va d i n h l y da d u a c c h u n g m i n h .
N h u v a y , cho t r u o c m o t tap bat k y , bao g i o c i i n g c6 the t h a n h lap m p t tap
CO l y c l u p n g I o n h o n . D o do, k h o n g c6 l u c l u p n g (ban so) nao la I o n nha't ca.
2.4.
'

Lyc LLTONG 2 " :

'^v'' '

-

••AM.A H ..n

mt,:

De ket t h i i c v a n de l u c l u p n g cac tap, ta hay t i m hieu t h e m ve t h i i bac tren

tap cac ban so.

.

;

.

. «

neu Bi va Bj la n h i i n g n g u d i khae n h a u , v | y cd m p t song a n h g i i i a n h i i n g
n g u d i quen A va B d o d d so n g u d i quen eiia A v a B n h u n h a u .
N e u D la m p t t r o n g n h i i n g n g u d i ed m a t t r o n g eupe h p p t h i hoac anh ta
quen v d i A hoac D k h o n g q u e n A t h i se cd 2 n g u d i q u e n c h u n g va k h i d o (theo
c h i i n g m i n h tren) A v a D cd n g u d i quen bang so n g u d i q u e n ciia A va cQng
bang so n g u d i q u e n ciia D . Vay A va D ed so n g u d i quen n h u n h a u . N h u vay,

Cho A la m p t tap bat k y , S la tap tat ca cac tap con (bp phan) cua A .

mpt n g u d i bat k y t r o n g n h i i n g n g u d i cd m a t d eupc h p p cd so n g u d i quen bang

N e u A la h u u h a n v a g o m n p h a n t u :

so n g u d i quen ciia A .

v o i m o i b p p h a n M ciia

,a2'-.^n

A ta CO the cho u n g day ( ^ i , ^ 2 ' - ' ^ n ) t^ong d o

= 0 neu 3 ; ^ M va

= 1 neu

Bj e M . Phep t u o n g u n g d o ro r a n g la 1 - 1 cho nen tap S t u o n g d u o n g v o i tap
tat ca cac d a y {^i.^Z'-'^n)

'^o the thanh lap d u p e bang n h i i n g soO v a 1 . N h u n g

de thay r a n g so day nay b a n g 2 " , v i m o i day g o m n so' c6 the lay hai gia t r i .

mtodn3.2.(TrungQuoc-1997).

Trong cac xau nhi phan c6 do dai n, goi a^, /« so cac xau khong chiea 3 so
lien tiep 0,1, 0 va \ so"cac xau khong chtta 4 so lien tiep 0, 0,1,1
0,0. Chihtgminh

rang:

Loi giai

M a r p n g k y h i e u do, ta q u i u o c m o t each t o n g quat rang: neu l u c l u p n g ciia

i-t

•(••••Vsv™:,*.!

Ta gpi m p t x a u thupe loai A neu no k h o n g chua 3 so' l i e n tiep 0, 1 , 0 va gpi

y i = 0 , y ^ = X i + X 2 + . . . + X i ^ _ i ( m o d 2). Rd rang X ehiia 3 so l i e n tie'p 0, 1 , 0 k h i va
chi k h i f(X) chua 4 so h a n g l i e n tiep 0, 0 , 1 , 1 hoac 1 , 1 , 0, 0 t i i c la X thupe loai A
^ ^ '

k h i va chi k h i f(X) thupe B.

K h o k h a n I a n nha't k h i giai m p t bai toan to h a p la xac d i n h h u a n g d i " N e u
CO m p t song a n h d i t u tap h u u h a n X d e n m p t tap h u u h a n Y t h i I X I = I Y I . T u
y t u o n g d o ta t i m d u p e l a i giai cac bai toan sau:

' *

Vay f la m p t song anh d i tir tap cac xau loai A d p d a i n d e n tap eac xau loai
B dp d a i n + 1 m a bat d a u bang 0. N h u n g t u m o i xau X thupe B ta n h a n dupe
mpt xau X Cling thupe B bang each d d i cac p h a n t i i ciia X theo q u y tac 1 - > 0, 0
~> 1 nen so cac xau loai B d p d a i n + 1 gap d o i so'cae xau loai B d p d a i n + 1 ma

/

Mgt cuoc hgp c6 n nguai: Mgt so nguai trong ho khong quen hiet nhau, dong
thai cH moi cap 2 nguai khong quen nhau lai c6 dung 2 nguai quen chung, con
hai nguai quen nhau lai khong c6 nguai quen chung. Chttng minh rang moi

mgt

con ngupi trong cugc hgp c6 mgt so nhu nhau nhung nguai quen.

Lin giai

Ta n h a n thay r a n g n h u n g n g u d i q u e n A t h i k h o n g q u e n n h a u (neu k h o n g
n h u v a y t h i h p c6 A la n g u a i q u e n chung). Gia s u B,

>,.-n. 4

0, 0. v d i m o i xau X = ( x i , X 2 , . . . , x „ ] , ta xay d u n g f(X) = ( y i . y a . - . y n + i ) n h u sau:

§3. C N G D U N G A N H X A D E G I A I

^di todn 3.1.

1,1,

mpt xau thupe loai B neu no k h o n g ehiia 4 so h a n g l i e n tiep 0, 0, 1 , 1 hoac 1 , 1 ,

A la a t h i l y c l u p n g ciia S (tap cac bp p h a n ciia A ) se d u p e k y h i ? u 2°'.

M O T SO BAI TO AN ROI R A G

hoac

= 2a^.

V g y S g o m 2" p h a n t u : I S I = 2" .
-

*

,

A j , A ^ (n > k) la

tap h p p tat ca n h i r n g n g u a i q u e n v o i A . K h i d o m o i m p t t r o n g cac A, k h o n g

bat d a u b a n g so'O. T i r d o ta ed d p e m .
N h g n xet: t u vi|e so sanh l u c l u p n g cae tap h p p , p h u a n g p h a p song anh cd
the g i i i p c h i i n g ta d e m so' p h a n t i i eiia m p t tap t h o n g qua s y so sanh l u c l u p n g
tap d d v d i m p t t^p khae ma ta da bie't so' p h a n t u eiia n o . ,»
^ d i todn 3.3. (Vd dich Ucraina Ggi M la so cac songuyen
^ong do CO n chit sol

1996)
duong viet

trong h? thap phan c6 2n chit so,

vd n chit so 2. ggi N la so tat cd cac soviet

phan CO n chit so, trong do chi c6 cac chU sol,
so 2. Chiing minh rhngM

= N.

2,3, 4 vd sochUsol

\'^{i'^^p]\m B i N H

trong h? thap
hang so chU
THUAN

17


Cty TNHHMIV

f H VHKhangVift

Lai giai
Vai moi so' c6 n chiJ' so' gom cac chu so 1, 2, 3, 4 va so' chu so 1 bang so' chu
so'2, ta "nhan d o i " thanh so'co 2n chu so'theo quy tac sau: dau tien, hai phien
ban cua so' nay duQc vie't ke nhau thanh so' c6 hai chu so', sau do cac chu so 3 6
n chu so'dau va cac chir s6'4 6 n chii so'sau dugc doi thanh chu sol, cac chu so
3 6 n chir so'sau va cac chir so 4 6 n chu so'dau dugc doi thanh chu so 2. v i du:
1234142
12341421234142 ^ 12121221221112.

^ai todn 3.5.
Chiing minh rang vai m, n, k e

N h u the^ ta thu dugic mot so' c6 diing n chii so' 1 va n chu so 2. R6 rang day
la mpt don anh. De chxmg minh day la mpt song anh, ta xay dung anh xa
ngugc nhu sau: voi moi so' c6 n chii so' I v a n chu so' 2, ta cat n chiJ so' dau va n
chii so'cuoi roi cpng chung theo cot voi quy tac: 1+1 = 1, 2 + 2 = 2 , 1 + 2 = 3, 2 + 1
= 4, va ta thu dugc mot so' c6 n chii so' gom cac chii so' 1, 2, 3, 4 vai so' chii so' 1
bang so cac so 2.

each chgn m + n + l - k s 6 ' t u m + n + l so nen se c6 C|^^n^.i each chgn. Cach thii

1212122

Vidu 12121221221112

1221112

'-m+n+l-^m*-n+^m-l'-n+l+-+'-m-k%+k

Loi giai
Ta

de'm

so

cac

l < a i 0 2 <-
bg

so

nguyen

T

=

vai

+ n + 1 bang hai each. Cach t h u nha't: ta xem do la so

hai: ta tha'y so cac bg T c6 a^^i_k =m + l - i la c|^":\C;,+i do do C^l^ each chgn
cac bg Ti=(ai,.-.,am-k) thoa lT2=(am+2-k am.n+i-k3 thoa m + 2-ichay tir 0 tai k

(do m + l-k
Chihtgminhrhng:


ih

f2^C^,6j;-_;^^/^^=C^2n.iyneZ'
k=0

'

>

'

'

,

!

'

Loi giai

>k.

Xetphep todn f dot vai bo sap thtctuX=

(a^.^z.-.a^+n+i-k)

^di todn 3.6. (Vd dick Trung Quoc -1994)

N h u the song anh giiia hai tap hop da dugc thiet lap va ta c6 M = N .
Cho cacso'nguyen duangn, k vain

,

c i ^ C ^ c i ^ T - i C i + "+CkCl^+k • Ket qua do cho ta dpcm.

1234142 n h u hinh ben.

1234142

(xi,...,x„) nhu sau: moi Idn chon k

so lien tiep tuy y trong X vd doi dau cua chung.

(1,1,1).

Lai giai
Xet bg thii t u X = ( x i , . . . , X n ) tuy y. Ta c6 2 nhan xet sau:

1. Co diing n - k + 1 nhom k so'lien tiep.
2. Sau mot so' chan Ian thuc hien phep toan f cho mot nhom k so' lien tiep
trong X thi gia tri k so' do khong doi.
N h u vay, moi phuong an thuc hien hiiu han Ian phep toan f cho X tuang
ung voi mot bg nhi phan A = (ai,a2,...,an), trong do ta tinh theo modun 2 cua
so Ian thuc hi?n f cho nhom k so'lien tiep (Xj.Xj+j,...,Xi+i^_i), va X tro thanh
(-lfi"""^'
(-l)a„_k.ix„)

Tu do, de tha'y moi bg A xac dinh duy nha't mgt bg X thoa dieu ki^n de bai
nen dap so'bai toan chinh la so cac bg A, tuc la 2"'"^^^.

va so' X. Sau do buac 1: ta chgn ra k cap roi t u moi cap chgn ra mgt so', buac 2:
le va khong dugc chgn ne'u n- k chan. R6 rang buoc 1 c6 2^C^ each chgn va

(i) Moi phdn ttt cua X deu thuoc tap {0,1}.
(ii) Co thethuc hien hitu han Idn phep todn fde ttt X nhan duac hg

Ta chgn ra n so t u 2n + 1 so nhu sau. Truoc het, t i i 2n + 1 so', ta chia ra n cap
chgn [(n-k)/2] cap trong n- k cap con lai, ngoai ra, so x se dugc chgn ne'u n- k

Tim so cac ho thii tuX = (x^,...,Xj^) thoa man cac dieu kien:

(i-iP^i

, m > k thi:

buac 2 CO c||"j^'^^^^^ each chgn va ta chgn dugc tong cgng n s6^ trong do k chay
tu 0 toi n. Lap luan do cho ta dpcm.

^di todn 3.7.
Cho truoc cac so nguyen duang m, n. Tinh T = ^ - ^ +
^
k=o 2""
k=o 2"
Lai giAi
Ta Chung minh tong can tinh bSng 1, tuc la:

Yj^),"^^+Yj^'^
k=0

•

k=0

Cac liiy thua cua 2 ggi ta lien tuang den so tap eon mgt tap hgp. );>
Trong cac tap eon cua tap S = {1, 2,
d^ng {ai,a2,...,a„+i},

m + n +1}, de tha'y c6

tap

( l < i < m + l ) trong do aj
V o i 0
n + k}, ngoai ra c6

* e "bo' sung" them mgt trong so 2'""'' tap eon cua tap (n + k + 1 , n + m +11.

^
19


Bdi duong hpc sink gioi Todn to hap - rdi r^c, Nguyen Van Thong

m

N h u vay

Cn+k 2"""

Cty TNHH MTV DWH Khang Vi$t

Lin giai

la so tap con ciia S c6 n h i e u h o n n p h a n t u .

D o f la m o t song a n h t u (1, 2 , n ) vao {1, 2 , n )

k=0

T u a n g hx

n
X'-'m+k^"'^ la so tap con ciia S c6 nhieu h o n m p h a n t u , cung

k=0

n

n Cm+k Z"""" la so tat ca cac t^p con cua S, tuc la 2'"+"+^.
V^y Xm Cn+k 2"'"'' + Z

f(l)

k=0
»t

I

n 1
l kI =-i f' W

Ta d u o c

Cho tap S gom tat cd cdc so"nguyen

trong doan [1; 2002]. Got T Id tap

hap

tat cd cdc tap con khong rong cua S. Vdi moi X thugc T, ki hi^u m(X) Id trung

Tu

(1)

va

(2)

ta

suy

ra:

binh cgng cdc phdn tit cua X. Tinh m = ^

Xay d u n g s o n g a n h f: T ^

• .8 6 • i t e J

'i 2^

5fa>^

T n h u s a u : f(X) = {2003 - x |x6X}, V X E T . R6 r a n g

(X) + m ( f ( X ) ) = 2 0 0 3 . D o d 6 :

^

n = l

a

'

2;Xni[X)=2(m(X) + ( f C X ) ) ) H T | . 2 0 0 3 ^ m = 2 ^ =

f(l)_f(2)_

^

vd cdc cha so"cua N thugc {1, 2,3,4,

2002 chit sothoa

man N:

5, 6, 7,8}.

G Q I M la tap cac so N thoa d i e u kien de bai.

I ^

f(n).

n2 V f M

"^(2)

' " '

f(n)

.•-.-^tls :>«ysi ivJ*::.::,, b

M

n

Vfrn
g'"'

N e u N = aia2..a2002

t

'

,

102002 _ i

- ,

^

xanh.

>

,

^

|i - k| c6 cdng nhau.

.,

Chicng minh td"t cd cdc phdn tic cua M CO cung mdu.

2002sfl^9

.

,

L a i giai
V o i m S i so n g u y e n a, dat Q so d u k h i chia a cho n v a k y h i ^ u Zn = {0, 1 , 2,

2002 so 9

•••/ n -1) t h i CO the coi p h e p to m a u da cho xac d i n h anh xa.

vd f: {1,2,..., n} ^ {1, 2 , n } Id mot song

^ >
- •
k=i k^ i ^ k

n,l
-1}.

2) Vai moi i cua M,i^k,ivd

N = X (N+f(N)3=|M|. 9 9 ^ .

ChOng minh rkng

< v ^ •,

4.1:(IM0-1985)

Ta gid sic:
1) VaimoiicuaM,ivdn-ic6cdngmdu.

2002 so 9

Suy ra t r u n g b i n h cpng cac s o ' N la : 99...9/2=

,;,(/>

Moi phdn tic ciia M dugc to bang mot trong hai mdu trang hodc

D o N + f ( N ) = 9 9 7 9 ;99 nen f la song anh M - » M .

<6di todn 4.0: Cho neH*

^ d i todn

tap hap {1, 2 , n

f ( N ) = bib2...b2oo2, v a i bj = 9 - Sj

NeA

99

n = l

Gid sic n Id mot sotyc nhien, k Id mot so nguyen to vai

,

Xay d y n g a n h xa f n h u sau:

20

„

(2)

^ k

m)

D a u bang xay ra <=> f ( i ) = i , V i = l , n

Loigiai

NeM

i ;

-I

1=1

Hay tinh trung binh cgng tat cd cdc so Ngom

2 X

^

k=l

1

^di todn 3.9:

Txxdo:

ktiVk^
— >

n

1

f(k)

k=l k

f(2)

(Tong l a y theo tat c a cac tap X thuQC T). '^'^

va

'V 2'

in

2002)

f(n)

f(2)

1'

D o c h i n h la d i e u phai chxmg m i n h . , , , i
todn 3.8. (VMO -

(1)

y J _ = ky= li'

A p d y n g bat d a n g thuc Cauchy - Schwarz cho 2 bp so.

tuc la so' tap con ciia S c6 k h o n g qua n p h a n t u .
k=0

n -1

1

dnh.

f:Zn

> Z2 = {trSng, xanh)

D i e u k i l n 1) cho i) f(a) = f ( - a ) , Va e Z
fa-

.

, ^

D i e u k i e n 2) v a i) cho ii) f(a) = f ( a ^ ) , Va e Z
21


Cty TNHH MTV DWH

Boi duang hgc sink gioi Todn to hap - rcri rac, Nguyen Van Thong

Khang Vi?t

Trong mgt so bai toan de'm, dinh l i tren thuang dugc ap dung bang each:

Vay vai mpi so nguyen m do ii) c6: f((m + l ) k ) = f[(m + k)k - k) = f(mk)

Ivluon de'm so phan tit cua tap hgp A , ta thiet lap mgt song anh tir A den B ma

V i (k, n) = 1 t?ip {mk/m € Z} = Zn

B la mgt tap hgp da biet so phan t u .

Vay f(Zn) la tap chi c6 1 phan t u .

Trong bai toan bat dSng thuc to hgp, y tuong tren thuang dugc ap dung

^ d i todn 4.2: Trong mgt hgi nghf todn hgc quoc tetochdc

tai quoc gia X, c6 cdc

nhu sau: Muon chung minh | A| < |B| , ta tim mgt anh xa d i tir A vao B (hoac B

nhd todn hgc trong mcac X vd nude ngodi. Moi nhd todn hgc quoc gia X gin

A) Sau do chung minh anh xa do la mgt dan anh (hoac toan anh) nhung

dung mgt thong di^ eho mgt nhd todn hgc nude ngodi. Moi nhd todn hgc nude

khong phai la song anh.

ngodi gid dung mgt thong diep cho mgt nhd todn hgc cua quoc gia X. Mac du

($di todn 4.3. ( I M O 1989). Mgt hodn vj (xi,X2,...X2n> cua tap hotp (1, 2,2n)

vay Cling ed it nhat mgt nhd todn hge khong nhan dugc mgt thong diep ndo cd.

Id so nguyen duang) dugc ggi la cd tinh chat P neu [xj - X j ^ i j = 1 vdi it nhat

Chihtg minh rang ton tai mgt tap hap S gdm cdc nhd todn hgc quoc gia X vd

mgt i e {1, 2 , 2 n

tap hap T cdc nhd todn hge nude ngodi thoa man cdc tinh chat sau:
nude ngodi khong thugc tap hap T.

(i) Cach 1. Dat A la tap tat ca hoan v i ciia (1, 2 , 2 n } ; B la tap tat ca hoan v i

2. Cdc nhd todn hgc nude ngodi thugc T chi gtti thong diep cho cdc nhd todn

ciia {1, 2,

hgc quoc gia X khong thugc S.
i •(-•">: ^ •

r.

1 Gpi A la tap hgp cac nha toan hpc cua quoc gia X va B la tap hgp cac nha
toan nuac ngoai. Goi f: A -> B va g: B

A la cac anh xa dinh nghia nhu sau:

f(a) la nha toan hpc nuoc ngoai nhan thong di^p ciia a va g(b) la nha toan hoc
cua quoc gia X nhan thong diep cua b. Neu cac tap con S va T ton tai thi T = B\
f(S). Vay ta phai chung minh ton tai mot tap S c A sao cho A\ = g(B\f(S)). Vai
m6i tap con X c A, goi h(X) = A\g(B\f(X)). Neu X c Y thi.
v;;f(x)cf(y)
.

;

'

2n} khong c6

tinh chat P.
Ta chia cac so 1, 2 , 2 n thanh n cap sau: (1; n +1), (2; n + 2 ) ; ( n ; 2n).
Gia su (x^; X 2 ; . . . ; x^^i; X;^; Xj^^^;...; X 2 n ) la mgt hoan v i thugc C.
Gia su

la so'ciing cap vai X 2 n , suy ra k < 2n - 2.

Ta xay dung anh xa f : C
X = (x^, X 2 ,

,

B nhu sau

X i ^ _ i , X|^, X | ^ ^ l , X 2 n ) -> Y = ( X j , X

2

,

,

,

>-• '^k+l )

De tha'y rang f la don anh va f khong phai la toan anh.

=>g(B\f(Y))cg(B\f(X))
=>A\g(B\f(X))c:A\g(B\f(Y))

That vay, xet yo = (x^, X2
x^-i • ^n+i • ^n+i
2n )
Khi do khong ton tai XQ de f( X g ) = YQ. D O do f khong phai la toan anh.

^ h(X) c h(Y)

T u do ta suy dieu can chung minh.

theo gia thiet g khong phai la toan anh vay ton tai HQ thugc A ma khong thugc
g(B\f(X)) vay thugc h(X) vai mgi X c A. Vay mgi tap hgp trong M chua ag.
Vay S = X e M n X khac rong. Theo dinh nghla cua S ta c6 h(S) c S. Theo tinh
chat dan d i f u cua h ta c6 h(h(S)) e h(S). Vay h(S) e M va S c h(S). Tong hgp
cac ket qua tren ta c6 S = h(S) (dpcm).

C| < n(2n - 1)!. T u do dan den each chung minh t h u 2 sau:
(ii)Cach 2. Ggi Aj^ la tap tat ca hoan v i thoa man k va k + n dung canh
nhau, A la tap hgp tat ca cac hoan v i c6 h'nh chat P. Suy ra A = (JAJ^ .
Theo nguyen ly bao ham loai tru, ta C O

Cho A va B la hai tap hCru han.
mgt dan anh f : A

k
B

thi |A| < |B| ;

2. Neu C O mgt song anh f : A -> B thi | A| = B| ;
3. Neu C O mgt toan anh f : A

B thi |A| > JB

Ta
'

> i M = n(2n-l)!

J^MnxetTa c6 |C|n|B| = 0 ma |C|+ |B| = |A| = (2n)! nen |B!

A=y:Ak-yKnAhK

Sir dung mgt so tinh chat ciia anh xa
CO

2nl c6 tinh chat P; C la tap tat ca hoan v j ciia {1, 2,

^B\f(Y)c=B\f(X)

G(?i M = {X c A/h(X) c X). Tap hgp M la khac rong v i da c6 A e M . Mat khac

1. Neu

- 1}. Chicng minh rang moi n, so hodn vi ed tinh chat P Idn

han so hodn vi khong cd tinh chat P.
Lai giai

1. Cdc nhd todn hge quoc gia X thugc S chi gid thu eho cdc nhd todn hgc

Loigiai

(n

CO

Suy ra

k
I

|AknAhnAj-...>SK|-IKnAh

k
= (2n-l)!,

k

A^nAhl = 4 ( 2 n - l ) !
nA,| = i

=2n(2r.-l)!,

k
,
^

•

4(2r.- 1)!

k•7^

99


Cty TNHH MTV DWH

Boi duang hoc sink gioi Todn to hap - roi rac, Nguyen Van Thong

Vi v^y

A| > 2n(2n - 1)! - ^^^^^^ • 4(2n - 2)! = 2n^ (2n - 2)! >

^

T u do ta C O dieu can chiing minh.
^di todn 4.4. (Romania TST 2002). Vai moi so nguyen ducmg, ta goi f(n) la so

each chon cac dau +, - trong bieu thiic:
= ± 1 ± 2 ±... ± n sao cho En = 0. Chiing
minh rang
(a) fin) = 0khin^l,2

(mod 4) ta c61

C O

the chi ra 6 truong hop khi xay dung

— ' — <<

f(n)

Do

do

anh

xa

tu

den

An

• T u do xac dinh

^"h.

An+4

Mat khac vol moi A n ta xac dinh dupe duy nha't mpt

, voi moi En cung

xac dinh dupe duy nha't mpt A n .
'

+ f(4k + 3) = f ( ( 4 k - l ) + 4 ) > 4 f ( 4 k - l ) > 4 ^ f ( 4 f - 5 ) > . . . > 4''f(3)

<2"-2L2J

(^)

= 4^2=-

Loi giai
(a) Gia su ton tai mpt each dat dau +, - voi n = 1, 2 (mod 4) de
Khi do 1 + 2 + ... + n = 0 (mod 2), suy ra "^"^^

An+4

dugc An la duy nha't.

Do do f(n + 4) > 4f(n). ta c6 f(3) = f(4) = 2. Suy ra

(mod 4).
+1

(b) Khi n=0,3

Do do ta

Khang Viet

= 0.

+ f(4k + 4) > f((4k - 1) + 4) > 4f(4k - 1) > 4^ f(4f - 5) > ... > 4'' f(4)

_(V2p

= 0 (mod 2). Ma dieu nay

xay ra khi va chi khi n = 0, 3 (mod 4). Ta c6 dieu can chiing minh.
(b) Ta chung minh f(n) < 2""^.
That vay, chia tat ca bieu thiic thanh 2""^ cap theo dang : (1 + 2a2 + Sa^ + ...+
n a ^ ; - ! + 2a2+ 3 3 3 + ...+ nan)v6i a; e (1;-1}, i = 2,n .
Neu f(n) > 2"~^ thi theo nguyen l i Dirichlet ton tai 2 bieu thuc cung nam
trong mot cap n tren, hieu cua chung bang 2. Do do chiing khong the dong thoi
bang 0 dugc (mau thuan).
^
> ^ »K '
+1

, .,»-

Do do f(n) < 2"-^ = 2" - 2""^ <2"-2^^^

j 4 f)

Ta chiing minh f(n) > -^^—'—.

1'

T u E„ ta se xay dvmg
*

Ta dinh nghia
•

,

A„+4

nhu sau

• Neu 1 € A n thi chon

An+4

= An

u {1; n + 1; n + 3);

•

Neu.2 e A n thi chpn

An+4

= An

\) u (n + 3; n + 4};

•

Nell 2 g A n thichon

An+4

= AnU{2;n

• Ta thay tap

An+4

,

* ^'

|BI| = |B2|.

.i

, B 2 Ian lupt la tap cac so'chan va le ciia B,
4

,

F :D ^ E

.

...

-^ "

Bh^Bin(Y\B2)

Ta c6:

|f(B)| =

n ( Y \I

= [B^| +

=n

|Y|-|B

Suy ra f(B) e E. De thay f la don anh. Ta chiing minh f la toan anh.

+ l ; n + 2).

N h u vay voi moi each chon A n ta c6 the xay dung it nha't 4 tap

An+4.

<5upc xay dung t u tap A n va

them diing mpt cap trong tap (n + 1; n + 2; n + 3; n + 4) va A n chiia diing mpt
cap nhu the'

• Tinh so tap can:

Xet anh xa
\n + 2 ; n + 4};

An+4

Gpi D la hp tat ea cac t^p can ciia A ; E la hp tat ca tap con n phan t i i ciia A .

khi do
= An

la tap tat ca cac so c h i n eiia A ; Y = {1; 3 ; 2 n - 1} la

Gia sii B la mpt tap can ciia A .

E„+4

An+4

bang sole. Bat so tap tot

2^""^

la T, so tap can la C. Chiing minh rang: T+C<
K i hieu X = (2; 4 ; 2 n )
tap tat ca eac so le ciia A .

^

N e u l e A n thichQn

Gia sir cho truac tap

tap tot neu trong tap do so cac sochan it han so cac sole. Mot tap con B cua A
duoc goi la mot tap can neu trong tap do sosochdn
Loi giai

(if

bang

(gdi todn 4.5. Cho tap hap A = {1; 2 ; 2 n } . Mot tap con C cua A duoc goi la mot

^jf't.'

Xet bieu thiic E n , gpi A n la tap hop cac so xuat hien trong E n vai dau + 6 truoc.
Neu En = 0 thi tong cac phan t i i cua

n) > — ' — . Vay ta c6 di
Vi vay f(n)
dieu can chiing minh.

That vay, gia sii M la mpt tap con c6 n phan t i i ciia A .
Gpi

,

IVI2

Ian lupt la tap cac so chan, cac so le ciia M .

Dat Bi = M l , B2 = Y \ M 2 , B = B^ u B2.

.

,

,

,

;
9=;


Cty TNHH MTV DWH

Boi duang hoc sink gioi Todn to hap - red rac, Nguyen Van Thong

B2I = |Y\M2| = |Y| - IM2I = n - IM2I = | M I | . D O d o B la

Suy ra

m p t tap can hay f la toan anh.

DAI s6 T 6 H O P

V i vay f la song anh. T u d o suy ra so tap can la C^n •
•

fj

T i n h so'tap to't:

§1.

Gia su C la m o t tap tot ciia A .

G p i F la hQ talt ca tap tot ciia A .

/'

tarn ky su. M o i k y t u c6 the la m p t c h i i so hay m p t c h u cai. M o i m a t k h a u p h a i

<J4t'}>^

chii'a i t nha't m p t chir so.

Ta thiet lap anh xa sau
G:E^F

H o i CO bao n h i e u m a t khau n h u vay? N h i r n g k y thuat can thiet de tra l o i

• ,,„:,;

cho cau h o i d o va m p t l o p r p n g Ian cac bai toan d e m khac se d u p e g i o i thieu

Cin(Y\C2)

trong c h u a n g nay.

Suy ra |g(C)| < n .
Ta CO g la m o t d o n anh nen |F| < C^n +

Bai toan d e m cac p h a n t u xuat h i e n rat n h i e u t r o n g toan hpc c u n g n h u t r o n g

+ ...+ C^'^ = 2^"-! - C 5 „ .

tin hpc. V i d u c h i i n g ta can d e m so n h i i n g ket qua t h a n h cong ciia cac t h i

T u cac dSng t h u c va bat d a n g t h i i c tren, ta c6 d i e u can c h u n g m i n h .
N h a n xet. Ta c6: C^n =

(2n)!

(n + l ) ( n + 2)...2n

(n!f

n!

nghiem va toan bp n h i r n g ket qua kha d i ciia n h i i n g t h i n g h i ^ m d o de xac d i n h
xac sua't cac bie'n so r a i rac. C h i i n g ta can t i n h so cac p h e p toan p h a i l a m t r o n g

>2L^J

mpt thuat toan de n g h i e n c i i u d p p h i i c tap ciia no.
T r o n g p h a n nay c h i i n g ta se t r i n h bay cac p h u o n g p h a p d e m co ban, c h i i n g

T u d o ta CO bai toan sau:
^di todn 4.6. Chiing

minh rang so'cac

thanh tong ciia cac songuyen
mot

so cho

truac)

xuai

him

thanh tong cac songuym

dumg
qua mot

each bieu dim

mot

ma khong c6 sonao
Idn khong Ian han

duang ma khong eo sonao

songuym

duong n

chia hei cho p (p la
so each bieu dim

chia hei cho

n

.

Lai giai
"

Xet bo f

, 8

2

,

}

^ao cho t r o n g bo k h o n g c6 p h a n t i i nao chia het cho p

la nen tang cho h a u n h u tat ca cac p h u o n g p h a p khac.
1.1.

Q U Y T A C

C O N G :

Chimg minh:

Xay d u n g song anh f : A - > (1,

=n

an} = { b i ,

b,,, p ^ i . t ^ ; . . . ; p'"'.tt)

G p i M la tap tat ca cac each bieu d i e n n t h a n h t o n g ciia cac so n g u y e n
d u a n g ma k h o n g c6 so nao chia het cho p (p la m o t so cho t r u o c ) xuat h i ^ n qua
m p t Ian va N la tap tat ca cac so bieu d i e n n t h a n h t o n g cac so n g u y e n d u o n g

a^} h->

bi,b2

"

'

'

• ''''

bi,,pti,pti,...pti,pt2,pt2,-pt2,-.Ptt>Ptf-Ptt
p-J^

so

p

De thay f la m p t d o n anh nen ta c6 d i e u can c h u n g m i n h .

'

+ n}

"^"^^^
neu X E B

,

1/

'1 111

i

1

V x i , x 2 e A w B , h ( x i ) = h(x2) v i do A n B = 0
nen f ( x i ) = f(x2) va g f x j + m ^ g ( x 2 ) + m

f ( x i ) = f ( x 2 ) v a g ( x i ) = g(x2)

^1 = ^2 ( V i f, g deu la song anh)

Vay anh xa h la d a n anh.

N
V-

{ai,

'

Ta c h i i n g m i n h h la song anh:

^

ma k h o n g c6 so' nao chia het cho p^.
Xet anh xa sau: f : M

v6ih(x)= j^W
[g(x)+m

'

2 , m }

h : A u B ^ { l , 2 , m

Bieu d i e n tat ca cac so chia het cho p d u a i d a n g p*" .t (r > 1 ; t / p).

•'r.-i
' V

Gia s i i |A| = m , IBI = n

g :B { 1 , 2 , n )

xuat h i ^ n qua m g t Ian va ^ ^ B j

82,...,

r

Cho A , B la hai tap him han va A n B = 0 t h i |A| + |B| = |A + B|

m

Giasu{ai,

PHfiPDEM

J\fhdn xet: M a t k h a u de t r u y cap vao m p t h | m a y t i n h g o m sau, bay hoac

G p i Ci, C 2 Ian l u g t la tap so chan va le ciia C t h i

Ch^

Khang Viet

*;A

•

N g o a i ra, h toan anh v i m p i p h a n t i i ciia h deu c6 anh va m p i p h a n t i i ciia
2 , m

+ n) d e u c6 tao anh nen h la toan anh. V a y h la song anh.

h : A u B -»{1, 2 , m

+ n} la song anh hay | A U B | = |A| + |B


Cty TNHH
Boi duang hqc sink gidi Todn to hap - rbi rac, Nguyen

1.2. QUY TAG C O N G GHO n TAP

1 4. QUY

HOP;

=0

t^pd6,thi

A;^ u . - . u A ^ = | A I | + | A 2 | + . . . + A ^ ] ,

neu i ; ^ j , t h i

Chung

Chung

minh

Gia su A^.-.A^ la cac tap h u u han d o i m p t r o i n h a u . Bang

TAG

N H A N G H O n TAP

| A I X A 2 X . . . X A „ | = |AI||A2|...|A„|

A i X...XAk XAk+i I = |(Ai X...XAk)xAk+i

= |(Ai X...XA
J^an

la k + 1 tap h i n i han d o i m o t r o i n h a u . K h i do

(Aj u...uAi^) va Aj^^i c i i n g r o i n h a u . Theo gia thiet q u i nap, ta c6:

=|(AiU...uAJuAk+i

xet qui tdc nhdn:

;

^

.

..^

.

Gia su m g t h a n h d g n g H bao g o m n giai doan ke tie'p va dgc lap v o i nhau,

+...+

+

Bj each thuc hien. K h i do hanh d g n g H c6 ca thay ai,a2,...,an each thuc hien.

iNhdn xet guy tdc cong:

'Bdi todn chia khda J.

Gia s u ta c6 n h a n h d g n g loai t r u Ian nhau Hi,...,Hn, t i i c la k h o n g the xay ra

Co bao nhieu anh xa t u m g t tap hgp X eo k phan t u t o i m g t tap Y c6 m phan t u .

hai h a n h d g n g d o n g t h o i . Ta c i i n g gia s u rang h a n h d g n g H i c6 3 ; each thuc

C O ca thay

Chicng

xay ra, hoac H2 xay r a , h o a c H„ xay ra,

h i f n. K h i do h a n h d g n g H : hoac

minh:

Giasu

Y = {yi,y2

«

A = {ai,a2,...,a^},
T i c h De-cac A x B

.

X

,

f(x)

B = {bi,b2,...,bk}.

g o m cac cap (aj.bj), 1< i < m, 1< j < k , c6 the v i e t thanh

m g t bang chir nhat c6 m d o n g k cgt n h u sau. /• 5
(ai-bi), (ai,b2)...(3i,bk)

'

( 3 2 , b i 3 , [a2,b23... ( 3 2 . b k )

I » J

>i Ji'i
'

i

u

X2
f(Xl)

D o n g t h u hai f ( x i ) , f ( x 2 ) ,

••

f(X2)

Xj

...

f(Xi)

Xk
f(Xk)

f ( X k ) la m g t day k p h a n t u cua Y. N o la

mgt p h a n t u ciia rich De-cac Y x Y x Y x

w/.,

/» ,t

y^}

M o i anh xa t u X t a i Y d u g c hoan toan xac d i n h b o i bang cac gia t r i cua no.

Cho A, B la hai tap h u u han va A n B = 0 t h i A x B = A . B
C h u n g m i n h : Gia s u |A| - m, |B| = k va

!

X = {xi,X2,.Mk}

+...+an each thuc hien.

1.3. Q U Y T A C N H A N :
'

|Ak+i|- |Ai| |A2|...|Ak||Ak+i

trong do giai doan t h u i la hanh d g n g H j . Ta cGng gia s u rang hanh d g n g Hj c6

=(AiU...uAJ
,j

,

Ta c i i n g c h i i n g m i n h dang thuc tren bang q u i nap theo n . V o i

minh

t u d i n h nghia t o n g cua hai ban so. Gia su dang thuc tren da d u g c c h u n g minh

AiU...uAkUAk+i

HOP.

...,Ak,Ak+i la k + 1 tap h i r u han bat ky. K h i do gia thiet q u i nap.

V a i n = 1, d a n g thuc tren hien nhien d i i n g . V o i n = 2, dang t h i i c tren suy ra

va Ai,...,A|^+i

Viet

tich cua h a i ban so'. Gia s u dang t h u c da d u g c c h u n g m i n h cho n = k > 2 va

q u i nap theo n, ta c h u n g m i n h rang: A^ u . . . u A n = A^ +...+ A„ .

cho n = k > 2

Khang

dang t h u c la hien nhien d u n g . V a i n = 2, dang t h u c suy ra t u d i n h nghia

6 day JAjj la luc l u g n g (so cac p h a n ttr) cua tap A j .
,

DWH

Neu A i , . " , A n la cac tap h u u han bat ly va AjX...xAn la tich De-cac cua cac

A j , A 2 , A n la cac tap h i i u han d o i m o t r o i nhau, t i i c la A j n A j

Neu

MTV

Van Thong

x Y = Y""

Dao l a i , m g i day k p h a n t u cua Y la (y„^ ,y^^ <-,yak)

xac d i n h m g t anh

'

x? f tir X t o i Y, neu ta dat f ( X i ) = y ^ j . Su t u o n g u n g d o g i i i a tap h g p cac anh xa
Cam-bi)' Cam,b2)...(a„,,bJ
D a t Aj = { ( a i , b i ) , ( a i , b 2 )
R6 rang |Aj| = k v o i l < i < m

^^
(ai,bk]}vdi
^

X t a i Y va rich De-cac Y"" la m g t song anh. Vay so anh xa riJ X t d i Y bang

yk

lI'v ^

ftu A j n A j = 0 n e u i ^ j va A x B = A i u A 2 U . . . u A n , . ! i > rUJ •
Theo q u i tac cgng, ta c6: |AXB| = |A]^| + IA2I +...+ |An,| = m.k = |A| . |B
Vay

= m

Jle qud:C6
Chung

bao n h i e u each p h a n p h o i k do vat vao m ngan keo ?

minh

M o i each p h a n p h o i la m g t anh xa tir tap h g p k d o vat vao tap

^liVp m ngan keo. V i v$y c6 m"^ each p h a n pho'i k do vat vao m ngan keo.

|AXB| = |A|.|B

29


Cty TNHH MTV DWH

Boi duong hoc sinh gioi Todn to hop - rcri rac, Nguyen Van Tfiong

1.5. C A C Vf DU MINH H O A .
Vi du l:Mot
Nguai

ChOng minh:

ban ddi c6 2 day ghe dot dien nhau, moi day gont c6 6 ghe.

ta muon xep cho ngoi cho 6 hoc sinh trudng A va 6 hoc sinh truang

B

trudng

ki 2 hgc sinh

ndo ngoi c^nh

nhau

hoac

dot di?n nhau

tht

khdc

cho ca hai tap h o p A va B tuc la A n B| d u o c t i n h hai Ian. D o d o
| A | + | B | = | A u B| + | A n B| <=> | A u B | = | A | + | B | - | A n B

t,) T n r t m g h
nhau.

A p d u n g cong thuc (1) ta co the t i m duoc so'phan t u cua ba tap h g p hxm han.

b) Bat ki 2 hgc sinh ndo ngoi dot di^
.''ii ;,

nhau thi khdc trudng

nhau.

Ta co: |A u B u C| = | A u (B u C ) | = | A | + | B U C| - | A n ( B U C ) |

L o i giai

D a n h so'cac ghe'theo h i n h ve
,

| A | + |B| la s o ' t i m duoc k h i ta de'm t r u a c he't so'tat ca cac phan

^ cua A r o i sau d o so tat ca cac phan t u B. N h u n g k h i do, so p h a n t u chung

vao ban noi tren. Hoi c6 bao nhieu each xep cho ngoi trong moi truang hap sau:
a) Bat

Khang Viet

®

®

®

Nhung A n ( B u C ) = ( A o B ) u ( A n C )

'' .

®

®

(6)

'

V ^ y |A n (B u C)| = |A n B| + |A n C| - |A n B n C|

V,;,;;,

Hay

®

® ® ® ® ®

a) Ghe
So' each xep cho

+ Ji^

1

2

3

4

5

6

7

8

9

10

11

12

12

6

5

5

4

4

3

3

2

2

1

1

IA

'

I

u B u C| = A| + | B | + |C| - 1 A n B| - 1 A n C| - |B n C|

qud:Va'\, B la tap

huu

han

bat

k y , ta l u o n co

' ''

+1A n B n C

A U B < [ A

+ B

D a n g t h u c xay ra <=> A n B = 0

+ Jie qud:Gia

six A , B la h a i tap hiJu han ne'u B c A t h i B

ngoi
Vidii 2:Trong

V a y so each xep 2 hoc sinh n g o i canh hoac d o i d i f n p h a i khac t r i r o n g la:

mot cdu veHinh

12 X 6 X 5^ X 4^ X 3^ X 2^ X 1^ =1036800
b) Ghe
So' each xep cho

mot dethi

1

12

2

11

3

10

4

9

5

8

6

7

12

6

10

5

8

4

6

3

4

2

2

1

tich,

hgc. Trong 60 tht sinh dvc thi, co 48 tht sinh gidi duoc cdu So

thi sinh gidi duoc cdu So hgc hoac Gidi tich, 50 thi sinh gidi duoc cdu Gidi

tich

hoac Hinh hgc, 25 thi sinh gidi duoc ca hai cdu So hgc va Hinh hgc, 15 thi

sinh

gidi duoc cd ba cdu. Hoi co bao nhieu thi sinh khonggidi

duoc cdu ndo?

Loi giai

V a y so'each xep 2 hoe sinh n g o i d o i d i f n phai khac la:

K i h i f u T la tap tat ca cac t h i sinh. A , B, C Ian l u g t la tap h g p cac t h i sinh

1 2 x 6 x 1 0 x 5 x 8 x 4 x 6 x 3 x 4 x 2 x 2 = 33177600

giai dugc cau So hoc, Giai tich, H i n h hgc. Theo t i n h chat 2 ta co:

Q U A : Q u i tac trir

AnB = A + B

A la m o t tap hxxu han, va B la m o t tap con cua A .
D a t B = A - B . Ta c6:

mot cdu veGidi

hgc, 40 thi sinh gidi duoc cdu Gidi tich, 32 thi sinh gidi duoc cdu Hinh hgc. Co 57

ngoi

1.6.

co ba caw. mot cdu veSohgc,

B n C =|B + C

B= A- B= A - B

A u B | = 48 + 40-57 = 31
B u C =40 + 32-50 = 22

I

n e n : A u B u C | = A | + | B | + |C| - | A n B| - | B n C| - |C n A | + | A n B n C|
Chung
CO:

minh : That v a y v i A = B u B va B n B = 0 nen theo q u i tac cpng ta
B

T u d o suy ra:
1.7. s6

= 48 + 40 + 32 - 31 - 22 - 25 + 15 = 57
V i (A u B u C) c T nen theo t i n h chat ta CO
T \ ( A u B u C ) | = | T | - | A u B u C | = 6 0 - 5 7 = 3.

= A - B

P H A N TCT C U A H O P H A I H O A C B A T A P H O P H O U

(1)

Vay CO 3 t h i sinh k h o n g giai d u g c cau nao.

HAN

a) Tnrcmg hgrp h a i tap h g p : Cho hai tap hop huxi han A va B ta c6 cong thuc.
A u B = A + B - A n B

' '

Vidu 3:Khi dim tra kei qud hgc tap cdc mon Todn, Ly, Hoa cua mot lap co
45 hgc sinh, nguai

ta nhan

thay: co 19 hgc sinh khong gidi mon ndo, 18 hgc

sink gidi Todn, 17 hgc sinh gidi Ly, 13 hgc sinh gidi Hoa, 10 hgc sinh gidi
^on

Todn vd Ly, 9 hgc sinh gidi hai mon Ly vd Hoa, 10 hgc sinh gidi hai

hai
mon

T^odn vd Hoa. Hdi bao nhieu hgc sinh gidi cd ba mon?

30

31


Boi duang hgc sinh gidi Todn to hop - rai rac, Nguyen Van Thong

Cty TNHHMTV DWH Khang Viet

Lai giai
Ki hieu T la tap hgp hoc sinh ciia lap. A, B, C Ian lugt la tap hgp cac hgc
sinh gioi Toan, Ly, Hoa cua lop do.
Vi A u B u C = T \ [ T \ ( A u B u C ) ]
Nen so'hoc sinh gioi it nhat mot mon la
t
A u B u C| = |T| - |T \A u B u C| = 46 - 19 = 26
Suy ra so'hoc sinh gioi ca ba mon la

• r
A n B n C = A u B u C - A - B - C + | A n B + B n C + Cr.B
= 26 - 18 - 17 - 13 + 10 + 9 + 10 = 7 ,, ,,
Vi du 4: Tim hieu kei qua hoc tap d mot lap hoc, nguai ta thay:
• Han - so hgc sinh dcit diem gioi a mon Todn cung dong thai dat diem
3
gioi a mon Vat ly;
; H
• Han - so'hgc sinh dat diem gioi & mon Vat ly cUng dong thod dat diem
3
gioi d mon Van;
^ thai dat dieM giai a
« Han -2 so hgc sinh dat diem gioi a Van cung dong
mon Lich sit;
e Han - so hgc sinh dat diem gioi a mon Lich sii cung dong thai dat diem
3
gioi a mon Todn;
Chicng minh rang trong lap hgc noi tren c6 it nhat mot hgc sinh dat diem
gioi cd ban mon Todn, Vat ly, Van vd Lich SM'.
(De thi MSG Quoc gia THPT Bang

B-2005).

Ldi giai
Ki hi|u T, L, V, S Ian lugt la tap hgp cac hoc sinh gioi Toan, Vat ly. Van,
Lichsu.
Theo
| T n Lde> bai,
- T , ta c6: L n V > (*)
V n S > 3— V

S n T > 3—
3
'
3
Ta giai bai toan bang phuang phap phan chung.
Gia su.khong c6 hgc sinh nao dat diem gioi 6 ca bo'n mon Toan, Vat ly. Van
va Lich su, khi do chi con: T n V = 0 hoac L n S = 0 .
Neu T n V = 0 thi ( T n L ) n ( L n V ) = 0 va ( T n S ) n ( S n V ) = 0 .
Ma ( T n L ) n ( L n V ) e L va
(TnS)n(SnV)cS
y'y

32

•(•If

Nen (TnL)+(LnV)|<|L|. Va |(TnS) + ( S o V ) <|S
Suyra |TnL| + |LnV| + |TnS| + |SnV|<|L| + |S| (1) , ; r:;;;. - :

Mit khac, tu (*) ta c6: |T n L| + |L n V| + |T n S| + |S n V| > | ( | T | + |L| + |v| + |s|)
•J

Ma |(|T| + |L| + |V| + |S|) = i[(|T| + |L|) + (|T| + |S|) + (|L| + |V|) + (|S| + |V|);
= i(|T u L| + |T n L| + | T u S| + |T n S| + |L u V| + |L n V| + |S u V| + |S n V|)
Nen
2(|TnL| + |LnV| + |VnS| + |SnT|)>|TuL| + |TuS| + |LuV|+ |SuV
,

>|L| + |S| + |L| + |S

Suy ra |TnL| + |LnV| + |TnS| + |SnV|>|L| + |S| (2)
Tix (1) va (2) ta gap mau thuan nen dieu gia su ban dau la sai.
• Neu L n S = 0 lap luan tuong tu ciing dan den dieu mau thuan. Bai toan
dugc chung minh ^
^, ,
.^j,, ,^^^.,
,. .
§2

T O H O P KHONG LAP

jVhan x^t: Gia su mgt dgi quan vgt c6 muai cau thu huan luy|n vien can
chgn nam nguai di thi dau a mgt truong khac. Ngoai ra ong ta cung chuan bi
mpt danh sach c6 thu tu gom bon cau thii de tham gia bo'n tran chai don.
Trong muc nay ta se nghien cuu cac phuang phap de'm so each chgn khong c6
thii tu nam cau thu de di thi dau va c6 danh sach khac nhau gom bon cau thii
tham gia tran choi don.
^
Tong quat han, chung ta se trinh bay cac phuang phap de'm so each chgn
khong CO thu tu cac phan tu khac nhau va nhirng sap xep c6 thii tu cac do'i
hrgng cua mgt tap hiiij han.

2.1. T6 HOP (J6 HOP KHONG LAP)

So'tap con ciia mgt tap hop c6 m phan tu.
Cho t^p hgp A c6 m phan tii. Ta hay xet xem tap hgp tat ca cac tap con ciia
^O/ T(A), CO bao nhieu p h a n tu.
^T(A) =2"
Chung minh:Ta dua ra hai each chung minh khac nhau.

33


Boi dtithi}; hoc sink gioi Todn to h
nong

CtyTNHHMTV

Cdch l . G o i Tg la tap hop tat ca cac tap con cua tap hop A chua phan tij
a 6 A . Hien nhien moi tap con nhu the dugc hoan toan xac dinh, neu ta biet ta't
ca cac phan t u con lai cua no (tru a). V i vay c6 bao nhieu tap con nhu the thi c6
bay nhieu tap con trong tap hop A ' = A - {a). Tap hop A ' nay c6 m - 1 phan tir,
Vi vay, neu ta gpi S^, la so tap con cua mgt tap hg-p c6 m phan tu, thi

Ggi Tg la tap hgp tat ca cac tap con ciia tap hgp A khong chua a, thi \g

rong- Mat khac ta cung c6 CJJJ = 1 , v i chi c6 mgt tap con chua m phan tir, do la
Chung rninh:
Muon dung mgt tap con k phan tir cua tap hgp A, ta c6 the ghep them vao
mpt t?P
^' ^ P^^^
^ "^9t trong m - ( k - l ) = m - k + l phan tir khong
thamduvaono.
Vi CO

T,uT,

tap con k - 1 phan tir va ta c6 the bo sung moi tap con ay thanh

mgt tap con k phan tir theo m - k + 1 each, nen lam nhu vay ta thu dugc
= 2s

Va Tg n Tg = 0 , nen theo quy tac cgng, ta c6: T ( A )

( m - k + l)Cm"^ tap con k phan tir cua A. Nhung khong phai tat ca cac tap con

m-l

nay deu khac nhau, vi ta c6 the dung moi tap con k phan tir theo k each, cu the

T u d o s u y ra s „ =2s„_i
Ap

dyng lien tiep dang thiic nay, ta dugc: S„ = 2Srn_i = 2^s^_2 = - -

la lay moi mgt trong k phan tir cua no ghep them vao k - 1 phan tir con lai. Vi
2'""^Si

Sj la tap con cua mot tap hgp c6 mot phan tu. Nhung mot tap hgp c6 mot
phan t u chi c6 hai tap con la toan bg tap hgp va tap hgp rong. Vay

= 2 . Do

vay so (m - k + l)ct^~^ vira tim dugc d tren gap k Ian so' cj^ cac tap con k phan
tir cua A. Do do ta c6 dang thuc: (m - k + l)cj^~^ = kCJ^

riji) i

Tirdosuyra:

d 6 : s „ = | T ( A ) | = 2'"

-s;

,k _ m - k + 1 k - i _
+
-•m
k
" ~
k
_ ( m - k + l)...(m-l) ,

Cdch 2: Phuang phap anh xa
Cho tap hgp A c6 m phan tir. Xet tap hgp Y = {0, 1). Voi moi tap con B cua
A , ta xac dinh mgt anh xa f : A -> Y nhu sau: Cho x e A , neu x e B thi ta dat f(x)

k(k-i)...2

= 1, con neu x g B thi ta dat f(x) = 0.

( m - k + 2) k_2 _
— —
k-1

™

la so tap con 1 phan tir cua A. No bang so phan tir ciia A, tuc la m. Vay

N h u vay ling voi moi tap con B ciia A , c6 mgt anh xa f tir A toi Y.

k

Dao lai, neu f la mgt anh xa tir A toi Y thi ung voi no c6 mgt tap con B cua
A , gom tat ca cac phan tir x e A sao cho f(x) = 1.
Svr tuong ung ay giira tap hgp cac tap hgp cua tap hgp A va tap hgp cac

m(m-l)...(m-k + l)
_ m ( m - l ) . . . ( m - k + l)(m-k)...3.2.1 _

m!

1.2...k(m-k)...3.2.1

k!(m-k)!

anh xa tir A toi X ro rang 1 - 1. Do do so tap con cua A , tuc la T { A ) , bMng so

.;
, „

Vi du 1: Cac duang cheo cua mot n -giac loi gap nhau tai bao nhieu diem,

anh xa tir tap hgp A c6 m phan tir toi tap hgp Y c6 2 phan tir. Ta bie't rang so
nay la 2™ . Vay:

Khang Vi$t

Chuy: Ta CO CjJ, = 1 , vi chi c6 mgt tap con khong chua phan tir nao, do la tap

bang sm-i vi cac tap con do ding la cac tap con cua tap hgp A ' = A - {a}.
Vi T{A) =

DWH

nhi hat ki 3 dubmgcheo nao cung khong cat nhau tai mgt diem?

|T(A)| = 2"

Loi giai

2.2. T 6 HOP:

Moi giao diem ling voi 4 dinh cua n -giac, dao lai 4 dinh cua n -giac xac

^inh nghia 2.2.1. Mgt tap con k phan tir cua mgt tap hgp m phan tir dugc ggi

•linh mgt giao diem (do la giao diem cac duong chte cua t i i giac xac djnh boi

la mgt to hgp chap k cua m phan tir.

cac dinh ay). V i vay so tat ca cac giao diem bang so to hgp chap 4 cua n dinh.

So' to hgp chap k cua m phan tir.

n(n-l)(n-2)(n-3)

m!

m(m-l)...(m-k + l)

k!(m-k)!

1.2...k

"
(1)

n(n-l)(n-2)(n-3)

1.2.3.4

24

Vi du 2: Co bao nhieu cdch phan phot 5 do vat cho 3 nguai, sao cho
^Sum deu nhan dugc it nhai mot do vat?

34

; f

jL

•

moi

, M
35

Boi duang hgc sitth gidi Todn to hap - riri rac, Nguyen Van Thong

Cty TNHH MTV DWH Khang Vi?t

L6i giai
N e u k h o n g ke d e n t i n h giao hoan ciia phep cpng, t h i so' 5 c6 the p h a n ticli
t h a n h m p t t o n g cua 3 so' ty; n h i e n khac 0 theo cac kieu khac n h a u sau day:
5=1+1+3=1+2+2=1+3+1=2+1+2=2+2+1=3+1+1.
L f n g v d i cac k i e u p h a n tich d o , c6 cac k i e u p h a n pho'i 5 d o v a t cho 3 nguoj
A , B, C sao cho m o i n g u o i d e u n h a n d u p e i t nhat m p t d o v a t n h u sau:
\ K i e u

II

III

IV

V

VI

A

1

1

1

2

2

3

B

1

2

3

1

2

1

C

3

2

1

2

1

1

Theo k i e u I , A n h a n d u p e 1 d o v ^ t t r o n g so'5 d o vat, v a y A c6 C3 = 5 each
chpn. B n h a n d u p e 1 d o v a t t r o n g so'4 d o v | t con l a i , v a y B c6 the chpn theo
C4 = 4 each. C o n C t h i n h a n 3 d o v a t con l a i theo d u n g 1 each. Theo q u y tac
nhan, so each p h a n p h o i theo k i e u I la C5C4.I = 5.4.1 = 20 each.
Lap

luan

tuong

t y , ta se tha'y

rang,

phan

phoi

theo

l^nnhat.

kieu

I I c6

C^C^.1 = 5.6.1 = 30 each; theo k i e u I I I c6 C^cil = 5.4.1 = 20 theo k i e u I V c6
2r2

C^C^.1 = 10.3.1 = 30 each, theo k i e u V c6 C^C^.l = 10.3.1 = 3 0 , theo k i e u V I c6

m-i
i

t u c la k = m = ^ . V a y so C2Jn la

" " ''^'^^^•

' • '

^'•

jsje'u n la le, n = 2 m + 1 t h i m < k < m + 1 , t u c la k = ^^—^ hoac k =
"

Z

Vay cac so C ^ + i =

I

Nguoi^-^

Mg'u n la chan, n = 2 m , t h i

la I o n nhat. ^ ^

.
z

,

Vi du 4: Cho da gidc dm AiA2...A2n (n^ N va n>2)

noi tiep trong

duang

trdn (O). Biei rang so"tarn gidc co dinh Id 3 trong 2n dinh Ai,A2,...,A2n nhieu
gap 20 Idn so hinh chit nhat co cdc dinh la 4 trong 2n dinh A^, A2, A 2 n . Tim n.
L o i giai
0'\

• So t a m giac tao t h a n h :

C3 =
2"

Z^")'

3!(2n-3)!

. i(2n)(2n - l)(2n - 2 )
6^

'

• V i da giac d e u va so' d i n h chan

''

nen so'cap d i e m d o i x u n g q u a t a m O la n .
Chpn 2 d i n h bat k i M , M ' d o i x u n g qua O
c o n each.

^. .) ,, ,.

„ .

,

,

,

- • ,'.:„

C h p n 1 d i n h N bat k i t r o n g cac d i n h con l a i co 2 n - 2 each.
L u o n l u o n t i m d u p e N ' d o i x u n g qua t a m O de M N M ' N ' la h i n h c h i i nhat.
N h u n g d o m o i h i n h chxx nhat M N M ' N ' n h u vay b i de'm t r i i n g l a i 4 Ian nen so

C^C^.1 = 10.2.1 = 20 each.
V i cac k i e u I , I I , I I I , I V , V , V I d o i m p t khac nhau, n e n theo q u y tac cpng, c6
ca thay 20 + 30 + 20 + 30 + 30 + 20 = 150 each.

Vi du. 3: Xdc dinh so Ion nhat trong cac so C°,

,

,c|;

u< u , - , .
, . , ,^ n { 2 n - 2 )
n{n-l)
hinh c h u nhat tao t h a n h la: -^^
=—
4
2

/ ,(

D o so' t a m giac n h i e u gap 20 Ian so' h i n h chCr nhat, nen:

.

Loi giai
Ta h a y so sanh h a i so'

ke n h a u .

o(2n-l)(2n-2) =30{n-l)

Vi r k _ n - k + l k-i

N e n Cj;-^ < C | ; , n e u

„

> C„, neu

o ( 2 n - l ) = 15<^n = 8
n-k +1
k
n-k +1

> 1 , tir d o k <
< 1 , t u do k >

1

1^1

"'fie.

+1

J

J ,

, do do

n-1

,
n+1

Vi du. 5: Mot tap theco 14 ngubi gom 6 nam va 8 nii trong do co An va Binh.

n +1

^guai ta muon chon 1 to cong tdc gom 6 ngubi. Tim so cdch chon trong moi

2

truang hap sau:

n +1

N e u Ci; la so I o n nhat, t h i cj;-^<c!; v a Cl;>c;;"^ t u d o k <
k + l>

(don>2)

«) Trong tophdi co mat cd nam Ian nit.
n +1

b') Trong tophdi

co 1 to truang, 5 to vim, han nita An vd Binh
j

^^oi CO mat trong to.

-

;1

*
L o i giai

6
14!
each chpn 6 n g u o i bat k i : €^4 = — = 3003
,
6!8

>;

m:.

•

khongdong

]
;"'"4

37


Boi duang hpc sink gioi Todn tohqp - rcri rac, Nguyen Van Thdng

Cty TNHH MTV DWH

So'each chon 6 nguoi toan nam: Cg = 1

Moi duang ngan nhat di tu diem (0, 0) den diem (m, n) deu gom m + n doan

8!
= 28
6!2!
Do do so' each chgn to eong tac de c6 nam Ian nvt. 3003 - (1 + 28) = 2974.
So' each chon 6 nguoi toan nu: Cg

b) Cdchl:

'^

'

'

"

"

+ IZC^^ = 4752

Tuong t u so'each chpn c6 Binh ma khong c6 A n cung la:

+ 12Cji = 4752

So each chon khong eo A n Ian Binh: 12C^i = 1 2 - ^ = 5544
Do do yeu cau bai toan: 2(0^2 + UC^i) + UC^^ = 2(4752) + 5544 = 15048

jigan nhat t u diem (0, 0) tai diem (m, n) bang so each ehpn n doan dpc t u m + n

c|^^_^)^,^ c6

y nghia hinh hpc nhu sau: Do la duang ngan

nhat tir diem (a 0) tai diem (m - k, k).

^ 'M " '

Chon tuy y 6 trong 14 hoc sinh eo:

Chii y: Ta eung eo the xet so' each ehpn m doan ngang thay cho n doan dpc.
Khi do so duong ngan nhat t u diem (0, 0) toi diem (m, n) se la Cj^^^ . N h u vay
ta da chung minh dupe bang hinh hpc dang thuc C^^^ = Cj^+n.
2.5. MOT S 6 TINH CHAT QUAN TRONG C U A C A C S6
2.5.i.Ne'u 0 < k < m t h i

Zi

.

(2)

Chicng minh:

each.

Chon A n va Binh roi chon them 4 hoc sinh trong 12 hoc sinh eon lai c6: C^2 each.

Vay so' each chon 6 hoe sinh do A n va Binh khong dong thai c6 mat:
Vol 6 hoc sinh da chpn xong c6 6 each ehpn ra to truong.
Vay so each ehpn thoa yeu cau de toan la:
= cf^.^j.^

Cdch 1: Theo y nghia hinh hpc cua so' cJ^=cL_|^wk, ta da thay rang
C(m-k)+k =

•

Cdch2:Neu

chinh la cong thuc (2)

dimg eong thuc (1) thi ta eo ngay

„m-k

sicf^ - ^ 1 2 ) = 15048 each.

m!
{m-k)!(m-(m-k)!)

m!
=C
(m-k)!k!

V nghia tap hop eua eong thuc (2) nhu sau: Gia su tap hop A c6 m phan tu.

•

Xet mot mang duang hinh chir nhat kich thuae m x n, tao thanh bai nhiing
hinh vuong, ngan each nhau boi n - 1 duong ngang va m - 1 duong dpc nhu
tren hinh ve. Ta hay tim so duong ngan nhat tren mang duang do de di t u goc
dual ben trai diem (0, 0) tai goc tren ben phai (diem (m, n)).
(0,n)-

Ne'u B mpt tap con k phan t u eiia A, phan bu B eua B trong A la mot tap con
(m - k) phan t u cua A. N h u vay giira tap hop cac tap con k phan t u ciia A va
tap cac con (m - k) phan t u eiia A c6 mot tuong ung 1 - 1. Do do so' tap eon k
phan tu bang so'tap con m - k phan tu. Dieu nay dupe dien ta boi eong thiie (2).
•2.5.2. Ne'u m va k la nhiing so t u nhien sao cho 1 < k < m - 1 thi ta c6.
pk _ p k - l

(m, n)

pi
-m-1

(4)

Cong thuc nay dupe gpi la quy tac Pascal
Chung minh:
£a£lLl: ViC;;-_\

•58

'

Cong thuc nay dupe gpi la quy tac do'i xung.

x ; , ,//

2,4. Y NGHlA HINH HOC CUA

thii t u ke'tie'p eiia eac doan ngang va cac doan dpc. Vi vay so tat ea cac duong

N h u vay, so Cj^ =

11!
. So'each chon A n lam to vien va khong c6 Binh: IZ.Cn - 12.-^^ = 3960
Vay so'each chon CO A n ma khong CO Binh:

thang, trong do c6 m doan ngang, n doan doe. Cac duang ay chi khae nhau boi

do^n dpc ngang, hie la bang Cj^+n .

= '

So each chon A n lam to truong va khong eo Binh: I.C12 =

Cdch2:

Khang Viet

(m, 0) X

Va C m - l

(m-1)!

(m-l)!k

(k-l)!(m-k)!

k!(m-k)!

(m-l)!

(m-l)!(m-k)

k!(m-l-k)!

k!(m-k)!

\

)


Cty TNHH MTV DWH Khang Viet

Boi dudmg hqc sink gioi Toan to hap - rcri rac, Nguyen Van Thong

_(m-l)!k

Nen

(m-l)!(m-k)

•""^ ~k!(m-k)!"^

k!(m-k)!

(m-l)!(k + m-k)

(m-l)!m

k!(m-k)!

k!(m-k)!

+... + c;;^ = 2"^

Vi da 6:Chiing minh hang dang thiicCl, + C^^ +
L6i giai

m!

Ta chiing minh rSng so tat ca cac tap con cua mot tap hop A c6 m phan t u

•=C
k!(m-k)!

bang 2"". Mat khac, neu goi k la so phan t u cua mot tap con cua A thi ta c6

Cdch 2;Cho tap hop A c6 m phan tu. Xet mot phan t u a € A. Khi do cac tap

0 < k < m . Vi so' tap con k phan tu ciia A bang

, va nhiing tap con c6 so

con k phan t u cua A chia thanh hai lop: lop thu nhat gom nhiing tap con

phan t u khac nhau khong giao nhau, nen theo quy tac cong, so tat ca cac tap

khong chua a, lop thii hai gom nhiing tap con chua a. Neu mot tap con k phan

con cua tap hop A bSng

tu khong chiia a thi no la mot tap con k phan t u cua tap hop A ' = A - {a}. Vi
vay so' tap con k phan t u ciia lop nhat bang so tap con k phan t u cua tap hgp m
- 1 phan t u A'. So' do la CJ^_i. Ta hay tim so' tap con k phan t u ciia lop thu hai.

Vay ta co: C + C^, + C^^ +... +

tap con cua lop thii hai bang so tap con k - 1 phan t u ciia tap hop A ' c6 m - 1
phan tu, tuc la

c]^"l\

Vi moi tap con k phan t u hoac chua a, hoac khong chua a nen no thupc hoac
lop thu nhat, hoac lop thu hai. Vay theo qui tac cpng, ta c6:

Cdch 3: So duong ngan nhat di tu diem O (0, 0) toi diem A (k, m- k) bang
k
pk
k+(m-k) ~ ^m-

=('^lf + ( c j , f +... +

•;

(qf

r. s^'S' •

Loi giai

So duong ngan nhat di tir diem O (0, 0) toi diem A (n, n) bang

. Moi

duong nhu the deu di qua mot va chi mot diem Ak (k, n - k) nam tren duong
cheo BD ciia hinh vuong ABOD.

,

So'duong di tir diem O toi diem Ak (k, n - k) bang c{^^^^_,^j =

. Mot duong

= k doan dpc. Vay so' duong di t u A k toi A bang c}^^^^_|jj =

. Theo quy tac

nhan, so duong ngan nhat di t u O toi A qua Ak bang CJ^.C^ ={Cn] • Theo quy
tac cpng, so'duong ngan nhat di t u O toi A bang tong so'duong di t u O qua Ak

Ai

m -k

Atk, m - kj
A2

i'

= 2^ (dpcm)

ngan nhat d i t u Ak (k, n - k) toi A (n, n) gom c6 n - k doan ngang va n - (n - k)

-m-1

<'

+ ci+... + .

Vi du 7: Chung minh hhng ddng thiic

Tat ca cac tap con nay deu chua a. Neu ta bo phan t u a di, thi ta duoc nhiing
tap con k - 1 phan tu, khong chua a va gom toan nhirng phan t u cua A'. Vay so

+ Cj„ + C^^ +...

: Ai'ffvjn ' i

if , (

(k = 0, 1,

n) (vi mot duong di qua mpt trong cac diem do thi khong di qua

cac diem khac).
Vilytac6C;„=(c«„f+(c;f-

. ' i l l

y
—•

0

Ta CO the chia cac duong do thanh hai lop khong giao nhau: lop thu nhat

A(n,n)

f:'.>n lotn m ?>

B
n-k

^

•••• ii'v--..-

Ak

k

^('k-l)+(m-k) - ^ m - l ' ^^P ^hu hai gom cac duong di qua diem A2 (k, m - k - 1),

so'phan t u cua lop nay la c{^^^j^_|^_^j = c[^_i. Theo quy tac cpng, ta c6:
pk _ / - k - l

pk

$:.

,

gom cac duong d i qua diem A^ (k - 1, m - k), so phan t u cua lop nay la
D

X

Vi diJL 8: Cho p la mot songuyen tole, n e N* (n > p ) . Tim so cac tap con A
cua tap {1, 2 , n ] co tinh chat sau.
1) A chiia diing p phan tic.
2) Tong cac phan tit cua A chia he't cho p.

^

jj,1

,


Boi dudng hoc sink gioi Todn to hap - red rac, Nguyen Van Thong

+ ... + X + 1

Xet da thuc p ( x ) = x P - ^ +

,

(1)

nen

XP

, a P ' ^ la p - 1 nghiem phan bi^t ciia P ( x ) va
- l =n(x-a')

,

,

Vidu 9:Cho bang vuong ktch thuoc: ^n^ + n + l j x ^ n ^ + n + l j

Nguoi ta to den S 6 vuong cua bang sao cho khong c6 4 6 to den nao tao

' D a thuc nay c6 p - 1 nghiem phuc phan biet. Goi a la nghiem bat ki cua p(x).
Vi a,

F

Cty TNHH MTV DWH Khang Viet

thanh 4 goc cua mot hinh chiinhat c6 canh song song vai canh cua bang vuong.
Chimgminhrang:

=1

S<{r\ \)><{n^+ +

Loi giai

iy:,'^^

Gpi a i ^ l < i < n ^ + n + l j la so'6 to den a hang thu i cua bang vuong. The thi
X e t d a t h i i c : Q ( x ) = f | ( x - a ' ) va goi H = { A c {1, 2,

n} : |A|=p} j ' j

n^+n+l

S=

3(6- hang thu i, so cac cap 6 den cung hang bang Cg. =

^

•Vay

i=0
G i a s u : Q(x) = ^ a V

,

'

!

n^+n+lg / g , _ j'\

tong socac cap 6 den cua bang vuong cung hang bang: A =

] ^
i=i

!
2

i Dlthaya„_p=-Xa^^^S(A)=X'^
So cac cap 6 c6 the cung hang bang: C^2
V i vay neu S ( A ) = i(modp) thi a^^^^ = a ' nen an_p

(2)

;

i=0

mot hang ngang thi khong c6 2 cap nao triing nhau.

,av ("'-^"F^"^^)>A^"'g^^a,(a,-l)

Mat khac, gia sir n = kp + r (0 < r < p-1).

Do vay

„, K h i d o : Q ( x ) = n ' ' ( x - a ' ) = ( x P - l ) n ( x - a ' )

Thanhthu

Do khong c6 4 6 to den tao thanh 4 goc cua mot hinh chii nhat c6 canh song
song voi canh bang vuong nen trong A cap 6 to den tren khi chieu xuong cimg

: O do so'ni la so'cac tap A e H ma S ( A ) = l(modp).

„

, .

i=l

(3)

i=l

i=0
' 111(3) riitra: an_p
-p =:k

^

n^+n+1

c>(n2 + n

n^ + n + l >

^X^i'l)-

i=l

(r r
: i

Y,^,a'=k
i=0

' 4

(4)

p-1

n +n+l

A2

Tic (4) suy ra a la mot nghiem cua R. V i degP = degQ v a a la 1 nghifm bat
Vay np_i = np_2 =... = H i = Ho = k

.|

p-1

a^-S

(1)

n +n+l

n'^+n+l

I

1^

n^+n+l

a? = (n2 + n +

l) X

a

i=l

i=l

i=l

T u ( l ) v a (2) ta c6:
S
-5

n^ + n + 1

ky ciia P nen P v a Q chi sai khac nhau hang so' nhan.

/

Z

i=l

i=0

^
i=l

Theo bat dang thuc Cauchy-Schwarz thi
S2 =

Xet da thiic R(x) = Plnjx'+ HQ - k

n^+n+1

S
^

>

^ '

Ini+no-k

p

;

'

Suyra: n o - k =J^i
P

«

P
CP

<^ S + n(n^ + n + l j

- k

Vay dap so cua bai toan la HQ = k + —

S - (n + l ) ( n ^ + n +1 j < 0

S < (n + l)(n^ + n +1

''n
d do k =

P

S - ( n + l ) ( n ^ + n + l j <0

P.

,1",

;

>' '

(2)


Cty TNHH MTV DWH Khang Viet

Boi dumtg hoc sink gidi Todn tohgrp - rat rac, Nguyen Van Thong

, §3. H O A N V I
N h a n xet: H o a n v i va c h i n h h g p

,

nen ?„, =xn?^_^ = m ( m - l ) P ^ _ 2 =... = m(m-l)...3.2.Pi = m ( m - l ) . . . 3 . 2 . 1 = m!

Uv,..„

- r ; ^ , i - o ' v / r - i b ftl

;v.;Er,,

H o a n v i cua m o t tap cac d o i t u g n g khac nhau la m o t each sap xep c6 t h i i t u

3.1.3. '6di todn chia khoa 2:C6 bao nhieu song dnh titmgt tap hop m phan tu:A
len chinh no?
«.v.i,,',v

cac d o i t u g n g nay. C h u n g ta ciing quan t a m t o i viec sap xep c6 t h u t u m g t so

Jiodnvi3.1.1:

'•

'

'

.

>

/i >

t u (tue la d u g c tao nen t u cixng m g t tap h g p ) d u g c g g i la n h u n g hoan v j cua

Sv t u a n g u n g g i i i a tap h g p cac song anh t u A len c h i n h n o va tap h g p cac
minh:

v i t r i theo m g t t h u t u xac d i n h .
6 v i t r i t h u nhat ta c6 the dat bat k i p h a n t u nao t r o n g m p h a n t u eiia A

(buoc 1 : m each chgn). Sau k h i da dat phan t u t h u nhat, 6 v i t r i t h u hai ta c6 the
dat bat k i p h a n t u nao t r o n g m - 1 phan t u con lai (buoc 2: m - 1 each chgn), v a
tiep tuc l a m n h u the eho t o i v i t r i t h i i m (buoc m : 1 each chgn). Theo

q u y tac

m ( m - l ) ( m - 2 ) . . . 2 . 1 = m!

,

J 7i.Kij g n i s b iMo'^^/lt

,

1.,. \

P„=m!

Cdch 2:Ta chgn m g t p h a n t u a nao do cua A , va xet tat ca cac hoan v i cua A
t r o n g d o a chiem v i t r i t h u nhat. H i e n nhien so hoan v i ay bang so hoan v j eiia
m - 1 p h a n t u eiia tap h g p A - {a}. V i vay so hoan v j ciia A t r o n g d o a chiem v i
t r i t h u nhat bang Pm - 1 . Ta k i hieu M la tap h g p tat ca cac h o a n v i cua tap h g p
A , Ma la tap h g p tat ca cac hoan v i ciia A t r o n g d o a chiem v i t r i t h u nhat, ne'u
A = {ai,a2,...,an,} t h i ta eo

t , ; . - n j U . -

M = Ma, u M , ^ U...UM
32

V i ro r a n g Mg. n Mg. = 0 ( i 5^
)
M
M

M
M

'32

ai

M

a2

. i f •

Vi di/. 1:C6 bao nhieu cdch sap thit tu t^p h
Chan deu c6 sohi^u chan?

Loigiii
T r o n g tap h g p {1, 2,

r J •

+

i'i"

i/--'

'

2nl c6 n so chan v a n so le. Ta c6 the dat cac so chan

vao cac v i t r i v o i so chan theo n ! each. V o i m o i each dat n h u t h e ' t u o n g u n g n !

each dat cac so' le v a o cac v i t r i c6 so h i f u le. V i vay theo q u y tac nhan, c6 ca
thay n!xn!=:(n!)^ each sap t h u t u tap h g p (1, 2,

nhan, so each sap t h u t u cac p h a n t u cua tap h g p A la

va VI

f la m g t a n h xa. A n h xa nay la d o n anh v i ne'u a^^ =A 3] t

hoan v i a i a A r o rang la 1-1. Vay so song anh tir A len c h i n h no bang P^ = m ! .

Cdch l:Ta se Ian l u g t chgn cac p h a n t u cua tap h g p A va sap c h i i n g vao m

Vay:

thi

m

a i a p h a n t u ak. V a y f la song anh.

1(3.1.2:

CO

Chung

•. •

f{3k) = ^ik "^^h '^'^(^l)- N ° ^^^S l3 toan anh v i p h a n t u ajj^ ( k = l,...m) la anh

tap h g p d o .

Ta

f(a^)} = A.

va v i f la toan a n h ne

Dao l a i eho m g t hoan v j a i a A :(ai^ .a^^ .....a^^). Ne'u dat f(ak)=ai|^ (k = l

N h u n g tap h g p sap t h u t u khac nhau, m a chi khac n h a u b o i t h u t u cac p h a n

^inh

A v i : d o f la d o n a n h , ta c6 f ( a i ) ; i f ( a j ) ( i 9 t j )
{f(ai),f(a2)

'

Loigiai •

Ne'u f la m g t song anh hi' A len A t h i f(ai),f(a2),...,f(an,) la m g t hoan v i ciia

p h a n t u d u g c g g i la m g t c h i n h h g p chap k cua n p h a n t u .

3,1.H0ANV!KH6NGLAP:

t.....v.:.

G i a s u A = {ai,a2,...,am}

p h a n t u cua m o t tap h o p . M g t each sap xep c6 t h u t u k p h a n t u eiia m g t tap n
K h i k = n , c h i n h h g p chap k ciia n p h a n t u c h i n h la hoan v i cua n p h a n t u

•

2n} thoa m a n y e u cau a i a

dau bai.

Vi dii. 2: Co bao nhieu hoan vi cua n phan tit, trong do 2 phan tic da cho
khdng diing canh nhau?
Loigiiii
T r u d c he't ta xac d i n h so' hoan v j , t r o n g d o hai phan t u a v a b da cho d u n g
c^nh n h a u . Co the xay ra cac t r u a n g h g p sau: a d u n g 6 v j t r i t h i i n - 1, b d u n g

<^ v j t r i t h i i n . S o ' t r u o n g h g p n h u the bang n - 1 . N g o a i ra a v a b c6 the d o i cho
cho n h a u d o d o c6 2(n - 1) each sap dat a va b canh n h a u . V o i m g i each do

h i o n g l i n g ( n - 2)! hoan v i a i a cac p h a n t i i khac. D o d o so' hoan v i t r o n g d o a

am

va b d u n g c^nh n h a u bang 2(n - 1) x (n - 2)! = 2(n - 1)!. V i vay so h o a n v i p h a i

= l , . . . , m ) , nen theo q u y tac cgng, ta c6.

timbang.

+ ...+
= P.m - l

n!-2(n-l)! = (n-l)!(n-2)

45


Cty TNHH MTV DWH

Boi dumtg hgc sink gidi Todn tohqp - roi rac, Nguyen V&n Thong

Vi du 3: Cho k chU hoa, m nguyen am, n phu dm (cd tat cdk + m+n chu). Tic
cdc chit do CO the lap duac baa nhieu "til" (mgt ti( la mot day chit cdi viet lien
tiep, khong nhat thiei phdi c6 nghia) sao cho trong moi tie do a vi tri dau phdi
la chic hoa, con ve cdc chic khdc phdi c6 r nguyen dm vd s phu dm trong so cdc
chicdd cho?
Loigiai
Truac het ta chon mot chu hoa. Co k each chon nhu the. Sau do tu" m
nguyen am ta chon ra r chu, dieu nay c6 the thuc hien theo

each. Cuo'i

cung tir n phu am, ta chon ra s chu, dieu nay c6 the thuc hien theo

each.

Khan^ v i e t

Thuanhan Po(0) = l .
£)e
Be ly giai quan h? (2) ta hay tinh so N cac cap (f, i) bang hai each:
Mpt mat, i chay qua k diem co dinh da xae djnh, nen moi hoan vi tt-ong
p„(k) hoan vi do co mat trong k cap (f, i). Boi vay N = k. Pn(k).
Mat khac neu f(i) = i, thi tren tap gom n - 1 phan tu con lai (tiie cae phan tu
Idiac i) hoan vi f eo k - 1 diem eo dinh, nen moi mot trong n phan tu i eo mat trong
p^_^ (k -1) cap. Do do N = n.Pn_i (k - 1 ) , nen dang thuc (2) duoe chung minh.
Tinh tong cac dang thuc d (2) theo k = 1, 2,
bang each thay n bang n - 1 ta eo:

Sau khi da chon nhu tren, ta dat chir hoa 6 v{ tri thu nhat, roi dat r + s chu con

n va dua vao dang thiic (1)
:
,|

|;kP„(k)=XnP„_i(k-l) = n ( n - l ) ! = n!

lai, dieu nay c6 the thuc hien theo (r + s)! each. Nhu vay theo quy tac nhan ta

k-O

k-1

.

lap duQC tat ea kC[^C^ (r + s)! tu thoa man cac yeu cau ciia dau bai.
Vi du 4: Cho m chit hoa A, B,Kvdn

chic thuang a, b, ...m, I. Tic cdc chic so

do CO the lap duac bao nhieu hodn vi sao cho moi hodn vi deu bat dau bang
mot cha hoa vd tan ciing hang mot chic thuang?
Loi giai
Lay mot cap chu: mot hoa mot thuong, chang han (A, a). Dat cac chu ay vao

§4.

Vidu 5:Cho tap S = {1, 2,n}

vain>l

vaifld mgthodn vi cua tap S. Phdn

So tat ca cac tap con k phan tu bang CJ^
Moi tap eon do co the duoe sap thu tu theo k! each.
Bang each ay ta tim duac tat ca cac tap con sap thu tu co k phan tu ciia tap
hgp A.
Vi vay so'tap con sap thii tu co k phan hi ciia tap hgp A bang k!Cj^ . So'nay

tic i cua S duac goi Id mot diem codinh neiifd) = i.

Ggi Pn (k) Id so hodn vi cua tap S c6 dung k diem codinh. Hay chtmg minh

''^'

Bay gio ta xet tap con sap thu tu cua mot tap hop m phan tu A. Ta xem tap
hg-p A nhu khong sap thu tu. Vi vay moi tap con cua no eo the duge sap thu tu
theo bat ky each nao.

tim duoe (m + n - 2)! Hoan vi thoa man dieu kien: bat dau bang A va tan cung
thay m.n.(m + n - 2)! hoan vi phai tim.

;

4 . 1 . TAP C O N SAP THU" TLJ C U A M O T TAP H O P :

vi tri cuo'i. Roi dat cac chu con lai theo du moi each vao gi&a hai chii' da lay, ta
bang a. Vi c6 the chon mn cap chu nhu cap (A, a), nen theo quy tac nhan, eo ca

C H I N H HOP

rn!

duge ki hi?u la A!1 . Ta CO: A!1 = ki•k!(m-k)!

m!

(m-k)!

'

Nhu vay ta da chiing minh dugc dinh ly sau:
k-O

4.2. C H I N H

Loi giai
1) Vi tong tat ca cac hoan vi eiia n phan tu eo 0,1, 2 , n diem codinh bang
tat ca cac hoan vi c6 the ciia n phan tu, hie bang n! nen eo dSng thuc.
i P n ( k ) = n!

(1)

k=0

'

,

2) Ta ehiing minh rang : Vk(l < k < n) deu c6
kF^(k) = n.F^_i(k-l)

(2)

HOP:

Cae tap con sap thii tu co k phan tii ciia mot tap hop eo m phan tii dugc ggi
la cac ehinh hgp chap k ciia m phan tii do.
Cac ehinh hgp chap k ciia m phan tii la khae nhau neu chiing eo nhiing
phan tii khac nhau, hoac neu eo nhiing phan tii nhu nhau, thi thii tu cac phan
do khac nhau.

©mh ly 4.2.1. Tap con sap thii tu eo k phan tii ciia mot tap hgp co m phan
hi bang.
A|^ = k!c!^

Diing (f, i) de ky hi^u cap gom hoan vj f tuy y cua n phan tu vol k diem co
dinh va i la diem tiay y trong k diem co'dinh do (tue f(i) = i).
iiillltiri

m!
= m(m-l)...(m-k + l)
(m-k)!

=

47


Boi duang hoc sink gioi Todn to hap - rcri rac, Nguyen Van Thong

Cty TNHH MTV DVVH Khang Vict

So'chinh hap khac nhau chap k cua m phan tit bang

.; fik^ = m ( m - l ) . . . ( m - k +1) =

Vidi^l:Chung minhrhng:

m!
(m-k)!

A^Jt^ + AjJtj^ = k^AjJ^k
Lai giai

A n+2

"n+k

= (n + k)!
=k^

^

.n+l

"n+k

Uk-2)!

((n + k ) - ( n + 2))!

((n + k ) - ( n + l ) ) !

;
Mpi chinh hpp chap k + 1 ciia n diem deu xac dinh mpt duong gap khiic ho

(k-1)!^

CO k canh trong mat phSng. Hai chinh hop khac nhau ciia ciing k + 1 diem, noi
chung, xac dinh hai duang gap khiic khac nhau (nhu tren hinh ve).

(n + k)!
,- ^

((n + k ) - n ) !

^n+k

Song v i cimg mpt duang gap khiic c6 the dupe cho theo hai each, ling v6i

^di todn chia kkoa 3: Tim so'dan dnh tic mgt tap Acok phan tit den mgt tap
B com phan tii (k < m).

hai chieu d i tren duong gap khiic do, nen trong tat ca cac chinh hpp chap k + 1
ciia n diem, moi duang gap khiic xac dinh bai mpt chinh hpp nhu the^ dupe
tinh 2 Ian.

Lai giai
Gia su A = {ai,a2,...,ak} va f la mpt don anh t u A tai B. f anh xa cac phan tu

,

Do do so'duang gap khiic ho phai tim la -AJ^^^ - i l i i l — — k )

ai,a2,...,ak cua A theo thu t u vao cac phan tu f(ai),f(a2),...,f(aij) cua B. V i f la

Trubnghap 2: Duong gap khiic kin.

dan anh nen aj ^^aj keo theo f ( a j ) 7 i f ( a j ) , (i 9^ j ; i, j = \

Moi chinh hpp chap k ciia n diem xac dinh mpt duang gap khiic kin.

k). Dgt f(ai) = bj eB

,

j.

<

thi {bi,b2,-,bk} la mpt chinh hgp chap k ciia m phan t u cua B. Vay mpi dan
anh f t u A tai B deu xac dinh mpt chinh hop chap k ciia m phan t u cua B.
Dao lai, gia su {b^,b2,...,bk} la mpt chinh hop chap k ciia m phan t u ciia B.
Neu ta dat f(ai) = bj(i = l,...,k) thi f ro rang la mpt anh x? t u A tai B. A n h xa
nay la don anh vi neu ajT^aj thi f(ai) = bj ^tbj = f(aj).
N h u vay so' don anh t u tap hop A c6 k phan tu toi tap hpp B c6 m phan t u
bang so'chinh hpp chap k ciia m phan tu, tiic la bSng AJ^ (k < m).
Vi du 2: Trong mat phang cho n diem, trong do khong c6 3 diem ndo thdng
hang. Co the dung dugc bao nhieu duanggap khiic ha hoac kin c6 k canh vd cd
dinh tai n diem dd cho?
Lai giai
Trubng hgtp 1. Duong gap khuc ho. Ne'u mpt duang gap khiic ho c6 k c^inh thi
so'dinh cua no bang k + 1 . V i vay ta phai CO n > k + 1.

..

^

, ,

Tir mpt chinh hop chap k ciia n diem, bang each hoan v i vong quanh, ta

dupe k chinh hpp khac nhau. Tat ca cac chinh hop nay deu xac d}nh cixng mpt
'lu'ang gap khiic kin.
Mat khac, cijng t u chinh hpp chap k ciia n diem tren day, neu ta hoan vj
vong quanh theo chieu ngupe lai, ta eung dupe k chinh hpp khac nhau nhung
'^ac dinh ciing mpt duong ga'p khiic ay.
N h u vay trong tat ca cac chinh hpp chap k ciia n diem, moi duang gap khiic
kin xac dinh boi mpt chinh hpp nhu the, dupe tinh k + k = 2k Ian.
Do do so duong gap khiic kin phai tim la

48

2k

A!^^^ =

2k

jO