Thiết kế bài giảng đại số và giải tích 11 nâng cao (tập 1) phần 2

Chifdtig: II
TO H d P VA XAC SUAT
Phan 1

^mifsG vA9f D £ CUA cm/dii^G
I. NOI DUNG
Ndi dung chfnh eua chuong II:
Quy tae ddm: Gidi thieu quy tac cgng va quy tac nhan va nhiing iing dung
cua cac quy tdc nay.
Hoan vi - ehinh hgp - td hgp : Day la ba quy tac ddm cu thd nhdm dd dem
cac phdn tii cua tap hgp hffu han theo cac quy luat thii tu ggi la hoan vi,
chinh hgp va td hgp.
Nhi thiic Niu-ton : Nhdm tim he sd ciia mdt khai tridn (a + b)".
Phep thit va bidn cd: Day la nhflng khai niem quan trgng cua xac sudt,
Trong bai cdn dua ra nhiing quy tdc tfnh xac su&.
Xae sudt cua cac bidn cd.
II. MUG TIEU
1. Kie'n thurc
Nam dugc toan bd kidn thiic co ban trong chuong da neu tren, cu thd :
Hinh thanh nhflng khai niem mdi cd hen quan ddn cac quy tac ddm.
Tfnh dugc sd cac td hgp, sd cac chinh hgp va sd cac hoan vi cua mdt taj
hgp gdm n phdn tit.
Phan biet dugc su khac nhau cua chinh hgp va td hop.
• xay dung dugc khdng gian mdu, each xac dinh bidn cd va xac suat.
118


2. KT nang
Suf dung thanh thao cdng thiic td hgp, chinh hgp va cac cdng thfle vd xae
suat.
• Ap dung tinh dugc cac bai toan cu thd.

3. Thai do
Tu giac, tfch cue, ddc lap va chu ddng phat hien cung nhu luih hdi kien
thfle trong qua trinh hoat ddng.
Cdn than, chfnh xac trong lap luan va tfnh toan.
Cam nhan duge thuc td eua toan hge, lihdt la ddi vdi xac sudt.
III. CAU TAO COA CHUONG
Ndi dung eua chuong gdm hai phdn du kidn duge thuc hien trong 21 tiet, phan
phdi cu thd nhu sau :
PM/iA.Tdhgp(8tidt)
§ 1. Hai quy tac ddm co ban

1 tidt

§2. Hoan vi, chinh hgp va td hgp

3 tidt

Luyen tap

2 tidt

§3. Nhi thfle Niu-ton

1 tidt

Luyen tap

1 tidt

Phdn B. Xac sua't (11 tidt)

§4. Bidn ed va xac sudt cua bien cd

2 tiet

Luyen.tap

1 tidt

§5. Cac quy tac tfnh xac sudt

2 tiet

Luyen tap

2 tidt

§6 Bidn ngdu nhien rdi rac

2 tidt

Luyen tap

2 tidt

6n tap va kiem tra chuong 2

2 tidt

119

Phan 2,
CAC B A I i§iOAIV
A. T 6

HOP

§1. H a i q u y t a c deiii cd b a n
(tiet 1)
I. MUC TIEU
1. Kien thurc
HS nam dugc:
• Hai quy tac ddm co ban : quy tac cdng va quy tac nhan.
• Bidt ap dung vao tiing bai toan : khi nao dung quy tdc cdng, khi nao diing
quy tdc nhan.
2. KT nang
Sau khi hge xong bai nay HS sfl dung quy tac dem thanh thao.
Tfnh chfnh xae sd phdn tfl ciia mdi tap hgp ma sdp xdp theo quy luat nao
dd (cgng hay nhan).
3. Thai do
Tu giac, tfch cue trong hge tap.
Bidt phan biet rd cac khai niem quy tac cdng, quy tdc nhan va van dung
trong tflng trudng hgp cu the.
Tu duy eac vdn dd cua toan hge mdt each Idgic va he thdng.
II. CHUAN Bj cClA GV VA HS
1. Chuan bj cua GV
- Chudn bi cac cau hdi ggi md.
- Chudn bi hinh 2.1.
Chudn bi phdn mau, va mdt sd dd dung khac.
2. Chuan bi cua HS

Cdn dn lai mdt sd kidn thfle da hge vd td hop d ldp dudi.
120


III. PHAN PHOI THOI Ll/ONG
Bai nay chia lam 1 tidt.
IV. TIEN TRINH DAY HOC
A. DAT VAN DE
Cdu hoi 1
Cd thd thanh lap dugc bao nhieu sd cd 3 chft sd khac nhau tfl cac chft sd 1,
2,3,4.
GV Cho HS liet ke.
Cdu hoi 2
Cho 10 chft sd, 0,1,...,9.
Cd thd liet ke dugc tdt ca cac sd lap tfl 10 chft sd tren dugre khdng?
GV: Ta thdy :
Rdt khd liet ke. Do dd phdi cd mgt quy tdc didim sd cdc phdn tit cua mdt
tap hgp.
B. BAIMCII

HOAT DONG 1

M6DAU
• GV neu bai toan trong SGK.
GV dat ra mdt vai cau hdi nhu sau:
?l|

Hay vidt mdt sd mat khdu.

GV chia ldp thanh 4 td, mdi td viet mdt so mat khdu, sau dd cho mdt ban

trinh bay xem cac td cd trung nhau khdng?
• Thuc hien [HIJ trong 3'
Hoqt dpng cda GV
Cdu hoi 1
Hay vidt mgt sd mat khdu.

Hoqt dpng cua HS
Ggi y tra Idi ckn hdi 1
Ir64j5, abcdeh, 123456,...

121


Hoqt dpng cda GV

Hoqt dpng cda HS

Cdu hoi 2

Ggi y tra ldi cau hdi 2

Cd the liet ke duge cap ki tu Khdng thd Uet ke trong mdt thdi gian
khdng?
nhdt dinh.
Cdu hoi 3

Ggi y tra ldi c^u hdi 3

Du doan sd m^t khdu?

Khdng du doan dugc.

HOAT DONG 2
1. Quy tac cdng
• GV neu va thuc hien vf du 1.
Hoqt dpng cda GV

Hoqt dpng cda HS
Ggi y tra ldi cau hdi 1

Cdu hoi 1

Cd bao nhieu each chgn tai Cd31 each chgn.
ldp 11 A?
Cdu hoi 2

Ggi y tra ldi ckn hdi 2

Cd bao nhieu each chgn tai Cd 22 each chon.
ldp 12B?
Cdu hoi 3

GcA y tra Idi c^u hdi 3

Tdt ca cd bao nhieu cadh
chgn.

31+ 22 = 53 each chon.

• GV neu khai niem quy tac cdng

Gid sit mdt cdng viec cd the dugc thuc hien theo phuang dn A hodc
phuang dn B. Cd n cdch thuc hien phuang dnAvdm cdch thuc hien
phuang dn B. Khi dd cdng viec cd the dugc thitc hien bdi n + m cdch.
Quy tac cdng bdi nhieu phucmg an
Gid sic mdt cdng viec cd the dugc thuc hien theo mdt trong k phuang
a« AJ, A2,..., A|j. Co nj each thuc hien phuang dnA^, n2 cdch
thue hien phuang dn A2,... vd n^ cdch thuc hien phuang dn A.^. Khi
do cdng viec cd the dugc thue hien bdi nj + n2 + • • • + n,j cdch.
122


GV thuc hien vi du 2. Vf du nay chi mang tfnh minh hoa.
• Thuc hidn [H2j trong 5'.
Muc dich. Kidm tra xem hge sinh da biet van dung quy tac cdng hay chua.
Hoqt dpng cda GV

Hoqt dpng cda HS
Ggi y tra ldi cau hdi 1

Cdu hoi 1
Cd bao nhieu de tai.
Cdu hoi 2

8 + 7 + 1 0 + 6 = 31 (each chgn).
Ggi y tra Idi cau hdi 2

GV ddi sd va hdi xem cd HStutraldi.
bao nhieu each chgn.

• GV neu each phat bieu khac cua quy tac cdng neu trong chu y.

Sdphdn tit cua tap hgp hitu hgn Xdugc ki hieu la |x| (hodc n(X)).
Quy tdc cdng cd the dugc phdt bieu dudi dgng sau :
Neu Ava Bid hai tap hgp hitu hgn khdng giao nhau thi sdphdn tilt
cua A uB bdng sdphdn tit cua A cdng vdi so phdn tif cua B, tUc Id
|AuB| = |A| + |B|.
Quy tdc cdng cd thd md rdng cho nhidu hanh ddng.
- Neu AJ, A2,..; A^ Id k tap hOu hgn vd A; n Aj = 0 vdi i ^ j
(vdii,j = 1,..., k) thi\AiKj A2^ ...yJ A^.\=\AI\

+ |A2| + " - + |A,^|.

- Hai tap hgp A, B bdt ki thi\A u B| = | A|+|B| - | A n B| .

HOAT DONG 2
2. Quy t^c nhan
• GV hudng ddn HS thuc hien vf du 3, sfl dung hinh 2.1.
Hoqt dpng cda GV
Cdu hoi 1

Hoqt dpng cua HS
Ggi y tra Idi cau hdi 1

Gia sfl tfl nha An ddn nha Cd 6.1 = 6 con dudng.
Binh cd 1 con dudng thi tfl
nha An ddn nha Cudng cd
bao nhieu each chgn?

123

Hoqt dpng cda GV
Cdu hoi 2

Hoqt dpng cda HS
Ggi y tra ldi c^u hdi 2

Hdi An cd bao nhieu each Cd 4. 6 = 24 each di tfl nha An qua nha
chgn dudng di den nha Binh ddn nha Cudng.
Cudng?
• GV neu quy tdc nhan
Gid sit mdt cdng viec ndo do bao gdm hai cdng dogn A vd B. Cdng
dogn A cd the ldm theo n cdch. Vdi mdi cdch thuc hien cdng dogn A
thi cdng dogn B cd the ldm theo m cdch. Khi do cdng viic cd the
thuc Men theo nm cdch.
f Thue hien |H3| trong 5'
Muc dich. Kidm tra xem hge sinh da biet van dung quy tac nhan hay ehua.
Hoqt dpng cda GV
Cdu hoi 1

Hoqt dpng cda HS
Ggi y tra ldi cau hdi 1

Mdi each dan nhan cd bao Viec lap mdt nhan ghd bao gdm 2 cdng
nhieu cdng doan, hay kd ten doan. Cdng doan thfl nhdt la chgn 1 chft
cae cdng doan dd.
cai trong 24 chft cai. Cdng doan thfl hai
la chgn 1 sd trong 25 sd nguyen duong
nhd hon 26.
Cdu hoi 2
Ggi y tra Idi cau hdi 2

Cd nhidu nhdt bao nhieu
Cd nhidu nhdt la 24.25 = 600 chide ghd
chide ghd dugc ghi nhan
dugc ghi nhan khac nhau.
khac nhau?
• GV cho HS md rdng quy tdc nhan cd nhidu hanh ddng.
Gid sit mdt cdng viec ndo dd bao gSm k cdng dognA^, A2,..., Aj^.
Cdng dogn Aj cd the thue hien theo Uj cdch, cdng dogn A2 cd
thi thuc hien theo n2 each,..., cdng dogn A^ cd the thuc hien theo
ny. cdch. Khi dd cdng viec cd the thuc hien theo nin2 •••ny. cdch.
124


• Thuc hien vf du 4
Hoqt dpng cda GV
Cdu hoi 1

Hoqt dpng cda HS
Ggi y tra ldi cau hdi 1

Mdi each lam mdt bidn sd xe Cd 6 cdng doan: Chgn 1 chft cai
may cd bao nhieu cdng doan, trong 26 chft cai; cdng doan 2 chgn 1
hay kd ten cac cdng doan dd:
chft sd, ed 9 each chgn, va 4 cdng
doan cdn lai mdi cdng doan chgn 1
chft sd va cd 10 each chgn.
>
Ggi y tra Idi c^u hdi 2
Cdu hoi 2

Thed quy tac nhan, ta cd tdt ca
Cd bao nhieu each lam mdt bidn
26. 9. 10. 10. 10. 10 = 2340000
sd xe may?
(bien sd xe).

• Thuc hien Vl du 5
Hoqt dpng cda GV
Cdu hoi 1

Hoqt ddng cda HS
Ggi y tra ldi cau hdi 1

Cd bao nhieu day gdm 6 kf ttt. Vl ed 26 + 10 = 36 each chon nen
mdi kf tu hoac la mdt chft cai theo quy tae nhan, ta cd the lap duge
(trong bang 26 chft cai) hoac la 36 day gdm 6 kf tu nhu vay.
mdt chft sd (trong 10 chft sd tfl
0 ddn 9)
Ggi y^tra ldi cau hdi 2
Cdu hoi 2
Cd bao nhieu day gdm 6 kf tu Vi mdi kf tu cd 26 each chgn nen
ndi d cau>a) khdng phai la mat theo quy tac nhan, sd day gdm 6 ki tur
khdng phai la mgt mat kh'du la 26
khdu?
Cdu hoi 3

Ggi y tra ldi eSu hdi 3

Cd thd lap duge nhidu nhdt bao
cd36^-26^

nhieu mat khdu?
125


HOATD6NG4

TOMTATBAIHQC
1. - Gia sfl mdt cdng viec co thd dugc thuc hien theo phuong an A hoac phuong
an fi. Cd n each thuc hien phuong an A va m each thue hien phuong an B. Khi
dd cdng viec cd thd dugc thuc hien bdi n + m each.
- Gia sft mdt cdng viec cd the dugc thuc hien theo mdt trong ^phuong
anAj, A2,..., Ajj. Cd Wj each thuc hien phuong anAj, n2 cdch thuc hien
phuong an A2,... va n^ each thue hien phuong an Aj^. Khi dd cdng viec ed
the dugc thuc hien bdi Uj + n2 + • • • + nj^ each.
2. - Gia sfl mgt cdng viec nao dd bao gdm hai cdng doan A va B. Cdng doan A
cd thd lam theo n each. Vdi mdi each thuc hien cdng doan A thi cdng doan B
cd the lam theo m each. Khi dd cdng viec cd thd thuc hi6n theo nm each.
- Gia sfl mdt cdng viec nao dd bao gdm k cdng doan Aj, A2,..., Aj^. Cdng
doan AJ cd thd thue hien theo Uj each, cdng doan A2 cd the thuc hien theo
n2 each,..., cdng doan Aj^ cd thd thuc hien theo n^ each. Khi dd cdng viec
cd the thuc hien theo njn2... Uj^ each.

BOAT DONG 3
Cdu 1.

Cdu 2.

126

MOT S6 CAU HOI TRAC NGHI£M

Mdt bai tap gdm 2 cau, hai cau nay cd cac each giai khdng lien quan
den nhau. Cau 1 cd 3 each giai, cau 2 cd 4 each giai. So each giai dd
thuc hien eac cau trong bai toan tren la tren la
(a) 3;
(b) 4;
(c)5;
(d)6.
Trd ldi. Chgn (c).
DQ giai mdt bai tap ta cdn phai giai hai bai tap nhd. Bai tap 1 cd 3
each giai, bai tap 2 cd 4 each giai. Sd cac each giai dd hoan thanh bai
tap tren la
(a) 3;
(b)4;
(c)5;
(d)6.
Trd ldi. Chgn (d).


Cdu 3.

Mdt Id hang duge chia thanh 4 phdn, mdi phdn duge chia vao 20 hop
khac nhau. Ngudi ta chgn 4 hop dd kidm tra chdt lugng.
Sd each chgn la
(a) 20.19.18.17;

(b) 20 + 19 +18 + 17;

(c) 80.79.78.77 ;

(d) 80 + 79 + 78 + 77.

Trd ldi. Chgn (e).
Cdu 4.

Cho cac chft sd: 1, 3, 5, 6,*8. Sd cac sd chan cd 3 chft sd khac nhau cd
duoc tfl cac sd tren la:
(a) 12;
(b) 24;
(e) 20;

(d) 40.

Trd ldi. Chgn (b).
Cdu 5.

Cho cae chft sd: 1, 3, 5, 6, 8. So cae sd chan cd 4 chft sd khac nhau cd
dugc tfl cac sd tren la:
(a) 4.3.2;

(b) 4 + 3+ 2;

(c) 2.4.3.2;

(d) 5.4.3.2.

Trd ldi. Chgn (c).
Cdu 6.

Cho cac chft sd: 1, 3, 5, 6, 8. Sd cac sd le cd 4 chfl sd khac nhau cd
dugc tfl cac sd tren la:

(a) 4.3.2;

(b)4 + 3+2;

(c) 3.4.3.2 ;

(d) 5.4.3.2.

Trd ldi. Chgn (e).
Cdu 7.

Mdt ldp hge cd 4 td, td 1 cd 8 ban, ba td cdn lai cd 9 ban.
a) Sd each chgn mgt ban lam 1 ^ trudng la
(a) 17;

(b) 35;

(c) 27;

(d) 9.

Trd ldi. Chgn (b).
b) Sd each chgn mdt ban lam ldp trudng sau dd chgn 2 ban ldp phd la
(a) 35. 34.32;

(b) 35+ 34+ 33;

(c) 35.34;

(d) 35.33.

Trd ldi. Chgn (a).
127


c) Sd each chgn 2 ban trong mdt td lam true nhat la
(a) 35. 34;

(b) 7.8 + 3.8.9;

(e) 35+34;

(d) 35.33.

Trd ldi. Chon (h).
HOAT DONG 6

HircJNG DAN B AI TAP SACH GIAO KHOA
Bdi 1
Hudng ddn. Sfl dung cac phuong phap ddm sd phdn tfl cfla mdt tap hgp.
Thep quy tac cdng, ta cd 5 + 4 = 9 each chgn ao so mi.
Bdi 2
Hudng ddn. Sfl dung quy tae nhan.
Chft sd hang chuc cd the chgn trong cac chft sd 2, 4, 6, 8; do do cd 4
each chgn. Chft sd hang don vi cd thd chgn trong eac chft sd 0, 2,4,6,
8; do dd cd 5 each chgn. Vay theo quy tac nhan, ta cd 4.5 = 20 sd co
hai chft sd ma hai chft sd cfla nd deu chan.
Bdi 3
Hudng ddn. Sfl dung quy tac nhan va quy tac cdng.
a) Theo quy tdc cgng, ta cd 280 + 325 = 605 (each chgn).

b) Theo quy tac nhan, ta cd 280.325 = 91000 (each chgn).
Bdi 4
Hudng ddn. Sfl dung quy tac nhan va quy tac cdng.
a) Cd 4.4.4.4 = 256 (sd cd bdn chft sd).
b) Ndu yeu cdu cae chft sd khac nhau thi cd 4.3.2.1 = 24 (sd).

128


§2. Hoan vi - Chmh hdp - To hdp
(tiet 2, 3, 4)
I. MUC TIEU
1. Kien thiirc
HS nam dugc:
Khai niem hoan vi, cdng thfle tfnh so hoan vi cua mdt tap hgp gdm n
phdn tfl.
HS cdn hidu dugc each chiing minh dinh If ve sd cac hoan vi.
Khai niem chinh hgp, cdng thfle tfnh sd cae ehinh hgp chap k cua n
phdn tfl.
• HS cdn hidu dugc each chihig minh dinh If vd sd cae chinh hgp chap k cua
n phdn tfl.
Khai niem td hgp, sd cac td hgp chap k cua n phdn tfl.
• HS cdn hidu dugc each chiing minh dinh li vd sd eac td hgp chap k cua n
phdn tfl.
HS phan biet dugc khai niem : Hoan vi, td hop va chinh hgp.
2. KT nang
Phan biet dugc td hgp va ehinh hgp bang each hidu sap xdp thfl tu va
khdng thfl tu.
Ap dung dugc eac cdng thfle tfnh sd eac chinh hgp, sd cac td hgp chap k
cua n phdn tfl, sd cae hoan vi.

Nam chac cac tfnh chdt cua td hgp va chinh hgp.
3. Thai do
Tu giac, tfch cue trong hge tap.
Bidt phan biet rd cac khai niem co ban va van dung trong tflng trudng hgp,
bai toan eu thd.
Tu duy cac vdn dd cua toan hge mgt each Idgic, thuc te va he thd'ng.

129


II. CHUAN B! CUA GV VA HS
1. Chuan bj cua GV
Chudn bi cae eau hdi ggi md.
Chudn bi phdn mau va mdt sd dd dung khac.
2. Chuan bj cua HS
• Cdn dn lai mdt sd kidn thfle da hge vd quy tac cdng va quy tdc nhan.
6n tap lai bai 1.
III. PHAN PHOI THOI LUONG
Bai nay chia lam 3 tiet:
Tiet 1: TU ddu den hit muc 2.
Tiet 2 : Tiep theo den hit muc 3.
Tiet 3 : Tiep theo den hit muc 4 vd bdi tap.
IV. TIEN TRINH DAY HOC
A. B A I

CU

Cdu hoi 1
Hay nhae lai quy tac cdng.
Cdu hoi 2

Hay nhae lai quy tac nhan
CduMoi 3
Phan biet quy tdc cgng va quy tdc nhan.
B. B A I Mdi
HOATDdNGl
1. Hoan vi
a) Hodn vi Id gi
• GV neu va hudng ddn HS thuc hien vf du 1.
GV cho HS didn va chd trdng theo each eua minh, sau Ao liet ke lai.
Giai

Cac kdt q uacd'thd

Nhdt
Nhi
Ba

130

, *


• Neu dinh nghia
Cho tdp hgp Acdn(n> 1) phdn tic. Khi sdp xip n phdn tie ndy theo
mdt thU tu, ta dugc mdt hodn vi cdc phdn tA cua tap A (ggi tdt la mdt
hodn vi cda A).
• Thue hien [HIJ trong 5'
Hoqt dpng cda HS

Hoqt dpng cda GV

Ggi y tra Idi c^u hdi 1

Cdu hoi 1
Hay kd mdt vai hoan vi.
Cdu hoi 2

GV cho HS kd va kdt luan.
Ggi y tra ldi c^u hdi 2

Hay kd tam hoan vi.

GVchoHSkd.

b) Sdcdc hodn vi
• GV neu vdn dd
?l|

Mdt tap hgp cd 1 phdn tfl cd bao nhieu hoan vi?

?2|

Mdt tap hop cd 2 phdn tfl cd bao nhieu hoan vi?

?3|

Mdt tap hgp cd 3 phdn tfl cd bao nhieu hoan vi?

• GVneu dinh ll 1:
Sdcdc hodn vi cua mdt tap hgp cd n phdn tii la

?^ = nt =

n(n-l)(n-2)...l.

• GV hudng ddn HS chiing minh dua vao quy tac nhan.
• GV ndu vf du 2, vi du nay chi mang tfnh minh hoa.
• ThiJfc hien |H2J trong 5'.
Hoqt dpng cda GV

Hoqt dpng cda HS

Ggi y tra Idi c^u hdi 1
Cdu hoi 1
Viec thanh lap cac sd cd la mdi viec lap sd la mdt hoan vi.
hoan vi khdng?

131


Hoqt dpng cda GV
Cdu hoi 2

Hoqt dpng cda HS
Ggi y tra ldi cdu hdi 2

Cd thd lap dugc bao nhieu Cd thd lap dugc 5! = 120 sd cd 5 chft sd
hoan vi.
khac nhau.
HOAT DONG 2
2. Chinh hgp

a) Chinh hgp la gi
• GV neu cau hdi:
Cho mdt tap hgp A gdm n phdn tfl. Viec chgn ra k phdn tfl dd sap xep ed thfl tu
?4|

Neu k = n, ta duge mdt sdp xdp ggi la gi?

?5|

Ndu k < n, ta dugc mdt sap xdp ggi la gi?

• GV neu vf du 3 va hudng ddn HS thuc hien.
• GV neu dinh nghia ,
Cho tap hgp A gom n phdn tic vd sd nguyen A vofl < k < n. Khi lay
ra k phdn tit cua A vd sdp xep chdng theo mdt thU tti, ta dugc mdt
chinh hgp chap k cda n phdn tit cda A (ggi tdt la mot chinh hgp
chdp k cua A).
?6

Hai chinh hgp khac nhau la gi?'

?7|

Chinh hgp khac hoan vi d didm nao?

• Thuc hien [H3J trong 5'
Hoqt dpng cda GV
Cdu hoi 1

Hoqt ddng cda HS

Ggi y tra Idi cau hdi 1

Liet ke so cac chinh hgp chap (a, b), (b, a), (a, c), (c, a), (b, c), (c, b).
2 cua 3 phdn tfl dd.
Cdu hoi 2
Cd bao nhieu chinh hgp?

132

Ggi y tra Idi cau hdi 2
Co 6 chinh hgp.


• GV neu nhan xdt:
Hai chinh hgp khdc nhau khi vd chi khi hodc cd it nhdt mdt phdn tic
cda chinh hgp ndy md khdng la phdri tic cua chinh hgp kia, hodc cdc
phdn tit cua hai chinh hop gidng nhau nhung dugc sdp xip theo thit
tu khdc nhau.
b) So cdc chinh hgp
• GV neu vf du 4 va cho HS thuc hien.
• GV neu dinh If
Kf hieu An la sd cae ehinh hgp chap k cua n phdn tfl (1 < k < n). Ta cd dinh If
sau day :
DjNH U
Sdcdc chinh hgp chdp k eua mdt tap hgp cd n phdn tit (1 A \ = n(n - l)(n - 2)...(n - k + 1).
GV hudng ddn HS chiing minh dua vao quy tac nhan.
• GV neu nhan xet trong SGK.
TU dinh nghia ta thdy mdt hodn vi eua tap hgp n phdn tit la mdt
ehinh hgp chap n cua tap dd nen A" = P^ = n!.

• GV neu vf du 5 cho HS thue hien. Co thd thay bdi vf du khac.
• GV neu ehu y tfong SGK.
Vdi 0
"

(n-k)!

Ta quy udc
Ol =

lvdAl=l.

• GV dUa ra cae eau hdi cung cd nhu sau:
Hay chgn dung sai ma em cho la hgp If.
?8|

Hoan vi n phdn tfl la chinh hgp chap n cua n
(a) Dung;

(b) Sai.
133


?9

Aj;iadungkhik>n.
(a) Dung;

?10

(b) Sai.

A„ la dung khi k < n.
(a) Dung;

?11

(b) Sai.

A„ = P„.

(a) Dung;

(b) Sai.
HOAT DONG 3

3. Td hgp
a) To hgp Id gi?
• GV neu dinh nghia.
Gid sit tap Aeon phdn tit (n >1). Mdi tap con gom k phdn tut cua A
ditgc ggi la mdt td hgp chdp k cua n phdn tiic dd cho
• Thuc hien |H4| trong 3'.
Hoqt dpng cda GV

Hoqt ddng cda HS

Cdu hoi 1

Ggi y tra ldi cau hdi 1

Liet ke eac td hgp chap 3 cua A.

{a,b,c}, {a,c,d}, {a,b, dj, {b,c,d}.

Cdu hoi 2

Ggi y tra ldi cau hdi 2

Cd bao nhieu td hgp?

Cd 4 td hgp

b) Sdcdc to hgp
• GV neu cac cau hdi:
?12|

Hai td hgp khac nhau la gi?

?13| Td hgp chap k cua n khac chinh hgp chap k cua n la gi?
• GV neu dinh li
Kf hieu Cn la sd cac td hgp chap k cua n phdn tfl (0 < k < n).
Ta cd dinh If sau day.
134


DINH Lf 3
Sdcdc tdhgp chap k cda mdt tap hoped nphdn tic(l
v_ A!: _n(n-l)(n-2)-(n-k + l)

" k!
ki
• GV hudng ddn HS chflng minh dinh li.
• GV hudng ddn HS thuc hien vi du 6 va vf du 7 nham cung cd kidn thfle vd
tdhgp.
HOATD0NG4
4. Hai tinh chdt cua C^
• GV neu tfnh chdt 1
/-ik _ p n - k

(0 $ Jk < n).

GV cd thd ehiing minh cho HS kha.
?14| Nhdc lai cdng thfle C„.
?15

n-k

TfnhC^

?16| Chiing minh cdng thfle tren.
• GV neu tfnh chdt 2.

c!;:}+c;_j=cj (!,• GV hudng ddn HS chflmg minh.
HOAT DONG 3

TOMTATBAIHQC
1. Cho tap hop A gdm n phdn tfl (n > 1).
Mdi kdt iqua cua su sap xdp thfl tu n phdn tfl cua tap hgp A dugc ggi la mdt

hodn vi cfla n phdn tfl dd.
Hai hoan vi cua n phdn tfl chi khdc nhau d thfl tu sap xdp.
P„ la sd cac hoan vi cua n phdn tfl. Ta cd
135


P„=n(n-l)...2.1.
2. Cho tap hgp A gdm n phdn tfl (« > 1).
Kdt qua cua viec ldy k phdn tfl khac nhau tfl n phdn tfl cua tap hgp A va sdp
xdp chflng theo mdt thfl tu nao dd duge ggi la mdt chinh hgp chdp k cda n
phdn tu: da cho.
An la sd eac chinh hgp chap k ciia n phdn tfl (l
Ta ed

Aj = n ( n - l ) . . . ( n - k + l ) .
3. Gia sfl tap A ed « phdn tfl (n > 1). Mdi tap eon gdm k phdn tfl eua A dugc ggi
la mdt td hop chdp k cua n phdn td da eho.
Cn la so cac td hgp chap k eua n phdn tfl (0 < kTacd
'"

k!(n-k)!

4. Tfnh chdt 1

Cl=Cl~^ .(0Tmh chdt 2 (cdng thfle Pa-xcan)

cS:!+ci;_j=cj(i<^<«).

HOAT DONG 6
MOT SO CAU HOI T R A C NGHlfiM

KHACH

QUAN

Hdy chgn khdng dinh ddng trong cdc khdng dinh sau, tic bdi 1 den bdi 4.
Cdu 1.

Cd 3 ban nam va 2 ban nfl sap vao 1 hang dgc.
a) Sd each sap xdp la :
(a)C^;

(b)Ci;

(c)5!;

(d)A3.

Trd ldi. Chgn (e).
b) Sd each sap xdp dd hai ban nft dflng hai ddu hang la :
136


(a) 3! + 2! = 8;

(b) 3!.2! = 12;

(c)5!;

(d)Ai.

Trd ldi. Chgn (b).
e) Sd each sap xdp dd hai ban nft diing kd nhau la :
(a) 3! + 2! = 8;

(b) 3!.2! = 12;

(c)2!x2!x3!;

(d)A^.

Trd ldi. Chgn (e).
d) Sd each sap xdp dd hai ban nam diing ke nhau la :
(a) 3!+ 2! = 8;
(c)2! x 2 ! x 3 ! ;

J

(b) 3! x 2! + 2! x 2! x 3!; = 12;
(d)A|.

Trd ldi. Chgn (b).
e) Sd each ldy ra 1 ban nam va 1 ban nfl la :
(a) 2;

(b)C§;

(c)5;

(d)3.

Trd ldi. Chon (c).
f) Sd each ldy ra 2 ban nam va 1 ban nft la :
(a) 2;

(b)Ci;

(c)5;

(d)3.

Trd ldi. Chgn (c).
g) Sd each ldy ra 1 ban nam va 1 ban nft la :
(a) 2;

(b)C|;

(c)5;

(d)3.

Trd ldi. Chgn (c).
Mgt ldp hge cd 20 ban nam va 15 ban nft.
a) Sd each ldy ra 4 ban nam va 4 ban nft di thi dau thd thao la :
(a)C^o;

(h)C\y,
137

(c)C^5+C^o;

(d)C^5.

Trd ldi. Chgn (c).
b) Sd each ldy ra 4 ban nam va 4 ban nft va mdt ban phuc vu di thi ddu
the thao la:
(a)Cf5+Clo+l;

(b)(Cf5+C^o).27;

(c)5!;

(d)C25+clo.

Trd ldi. Chgn (b).
e) Sd each ldy ra 3 ban nam va 4 ban nft va mdt ban phuc vu di thi ddu
thd thao la:
(a) Cts +clo+ 1;

(b) (Cf^ +cio). 27;

(e)(Cf5+Cio).28;

(d)Cf5+C^o-

Trd ldi. Chgn (c).
Cdu 3.

Sd cac sd cd 3 chft sd khac nhau ma chft sd tan cung la 2 hoac 5 la:
(a)Afo;

(b)A^;

(c)Ai;

(d)2(Al-Al).

Trd ldi. Chgn (d).
Cdu 4.

Sd cac sd cd 4 chft sd khac nhau khdng chia hdt eho 10:
(a)Afo-2A5;

(b)Afo-A5;

(c)Afo;

(d)A^

Trd ldi. Chgn (a).
Cd^ 5.

138

Hay didn dflng, sai vao d trdng cua nhiing khang dinh sau:
(a) Sd each chgn 4 trong 7 ngudi di du hdi nghi la A7

[J

(b) Chgn 4 trong 7 ngudi di du hdi nghi la C7

[J

(c)C^=35

D


D

(d)A5=840
(a)

(b)

s

S

(c)
D

(d)
S

HOAT DONG 9
HLfCnSlG D A N B A I T A P SGK

Bdi 5
Hudng ddn. Sfl dung kidn thfle ve hoan vi.
Cd5! = 120 kha nang.
Bdi 6
Hucmg ddn. Dua vao chinh hgp.
Cd

A|

= 8.7.6 = 336 ket qua ed the.

Bdi 7
Hudng ddn. Sd doan thang la sd cac td hgp.
Sd cae vecto la sd cac chinh hgp.
a) vay sd doan thang ma hai ddu mflt la hai didm thudc P chfnh bang
,2 =
_ n(n-l)
sd td hgp chap 2 cua n phdn tfl, tfle la bang C^
b) Sd vecto cdn tun bang sd chinh hop chap 2 cua n phdn tfl, tfle la
bang An = n(n-\).
Bdi 8
Hudng ddn. Khdng phan biet chfle vu thi dp dung td hgp.
Phan biet chfle vu thi sfl dung chinh hgp.
a) Cd C7 = 35 each chgn.
b)Cd A7 = 210 each chgn.

139


Luyen tap (tiet 5, 6)

I. MUC TIEU
1. Kien thurc
HS dn tap lai
Quy tdc cgng va quy tae nhan.
Khai niem, cdng thfle tfnh so cac td hgp, ehinh hgp hoan vi.
HS phan biet dugc khai niem : Hoan vi, td hgp va chinh hgp.
2. KTnang
Phan biet dugc td hop va chinh hgp bang each hidu sdp xdp thfl tu va
khdng thfl tu.
Ap dung duge eac cdng thfle tfnh sd cdc ehinh hgp, so cac td hgp chap k
eua n phdn tfl, sd cac hoan vi.
Ndm chdc cae tfnh ch& cua td hgp va chinh hgp.
3. Thai do
TU giac, tich cue trong hge tap.
Bidt phan biet rd cac khai niem co ban va van dung trong tflng trudng hgp,
bai toan eu the.
TU duy eac vdn dd cua toan hge mdt each Idgic, thuc td va he thdng.
II. CHUAN Bj COA GV VA HS
1. Chuan bj cua GV
Chudn bi cac cau hdi ggi md.
Chudn bi phdn mau va mdt sd dd dung khac.
2. Chuan bi cua HS
Cdn dn lai mgt so kidn thfle da hge d bai 1 va bai 2.
III. PHAN PHOI THOI LUONG
Bai nay chia lam 2 tidt:
140


IV. TIEN TRINH DAY HOC

A. BAI CU
Cdu hoi 1
Neu cdng thfle tfnh sd eac td hgp, ehinh hop, hoan vi cua tap hgp gdm n
phdn tfl.
Cdu hoi 2
Phan biet td hgp, chinh hgp.
Cdu hoi 3
Neu cae tfnh chdt eua td hgp.

B. BAI Mdi
HOATDONGl
Bdi 9
Hoqt dpng cda GV

Hoqt dpng cda HS

Cdu hoi 1

Ggi y tra Idi c^u hdi 1

Gia sfl ed mdt eau trac Cd 4 phuong ai.
nghiem, hdi cd mdy phuong
an?
Ggi y tra Idi cau hdi 2
Cdu hoi 2
Bai thi cd 2 cau thi cd bao Cd 4. 4 = 4^ phuong an
nhieu phuomg an?
Ggi y tra ldi cau hdi 3
Cdu hoi 3
Bai thi cd 10 cau thi cd bao Cd 4^° = 1048576 phuong an tra ldi.

nhieu phuong an?
HOAT DONG 2
Bdi 10
Hoqt dpng cda GV
Cdu hoi 1
Hay lap mdt sd cd 6 chfl sd.

Hoqt dpng cda HS
Ggi y tra Idi cau hdi 1
ahcdeg.

Cdu hoi 2
141


Hoqt ddng cda GV

Hoqt dpng cda HS

Cd mdy each chgn g.

Ggi y tra ldi cau hdi 2

Cdu hoi 3

Cd 2 each chgn g la : 0 hoac 5.

Cd may each chgn a?-

Ggi y tra Idi cau hdi 3

Cdu hoi 4

a e (.1, 2,..., 9}, cd 9 each chgn.

Cd mdy each chgn b, c,d,e?

G0i y tra ldi cau hdi 4

Cdu hoi 5

b, c, d, ee {0, 1,..., 9}, mdi sc>' cd 10
Sd cac so cdn tim la bao each chgn.
nhieu?
Ggi y tra Idi cau hdi 5
Cd 9.10.10.10.10.2 = 180000 sd nhu
vay
HOAT DONG 3

Bdi 11
Hoqt dpng cda GV

Hoqt dpng cda HS

Cdu hoi 1

Ggi y tra ldi cau hdi 1

Cd bao nhieu phuong an di tfl Cd 4 phuong an:
AddnG.

\)A^B^D

^E^G;

2)A'> B ^ D ^ F ^ G ;
3)A-» C ^ D ^ E -^ G;
4)A -^C

^D^F^G.

Ggi y tra Idi cau hdi 2

Cdu hoi 2

Mdi phuong an tren cd bao GV chia HS lam 4 td, mdi td lam mdt
cau. Dua vao quy tac nhan.
nhieu each di?
Ggi y tra ldi cAu hdi 3
Cdu hoi 3
Tdng cdng cd bao nhieu Cdng 4 phuong an tren lai.
phuong an?
,

142

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