Tài Liệu Các Chuyên Đề Toán Học ( Cực chất) tổ hợp hay - MỘT SỐ CHUYÊN ĐỀ VỀ TỔ HỢP DÀNH CHO HỌC SINH CÓ NĂNG KHIẾU TOÁN BẬC TRUNG HỌC PHỔ THÔNG

Một số chuyên đề về tổ hợp dành
cho học sinh có năng khiếu toán
bậc trung học phổ thông


1
2
3
4
5
6
7
8
9
10
E
i
(i = 1, , k) k
(i)
(ii) E
i
n
i
k (n
1
+ n
2

+ + n
k
)
18 12
18 + 12 = 30
E 10 F
10
E 4 F 4 2
E F 4 + 4 −1 = 7
E
i
(i = 1, , k) k E
1
n
1
E
2
n
2
E
1
E
3
n
3
E
1
E
2
E

k
n
k
k −1 k
n
1
.n
2
.n
3
n
k
6 8
10
(i) 6.8.10 = 480 3
(ii) 6 + 8 + 10 = 24 1
8 3
8 3
8
= 6561
X n r
n
r X r
n X
n X X
r n P (n, r)
{2, 3, 4} {2, 4, 3} 3 X =
{1, 2, 3, 4, 5}
r X r X
r n C(n, r)

(i) P (n, r) =
n!
(n − r)!
(ii) C(n, r) =
P (n, r)
r!
=
n!
r!(n −r)!
= C(n, n − r)
m! ≡ (1).(2) (m) 0! ≡ 1
(i) n X
r (n − 1) (n − 1)
r
n(n − 1)
r
P (n, r) = n(n −1) (n −r + 1) =
n!
(n − r)!
(ii) C(n, r) r n X
r X
r r
P (n, r) = P(r, r) + P(r, r) + + P (r, r)
r X C(n, r)
P (n, r) = C(n, r)P(r, r) = C(n, r)r!
r X (n − r)
C(n, r) = C(n, n − r)
n
P (n, n) = n!
r

n r n r
n r n
12 12 10 11
9 10 4 12
4 11 3 10 C(12, 4) =
12!
4!8!
= 495
4 12 C(10, 4) = 210 4 11
C(9, 3) = 84 3 10
495.210.84 = 8731800
n (n + 1)
2
n = 3
n + 1 = 4
n kn + 1
k k + 1
1.3.1 6
n = 3
k + 1 = 6
kn + 1 = 16 16
20 4 7
9
r = 4, 5, 6, 7, 8, 9
∗) 1 r = 4 = k + 1 k = 3 3 n = 3
kn + 1 = 3.3 + 1 = 10
∗) 2 r = 5 = k + 1 k = 4
4
r = 5 n = 2
kn + 1 = 4.2 + 1 = 9

4 + 9 = 13
∗) 3 r = 6 = k + 1 k = 5
2 4 + kn + 1 = 4 + 5.2 + 1 = 15
∗) 4 r = 7 = k + 1 k = 6
4 + kn + 1 = 4 + 6.2 + 1 = 17
∗) 5 r = 8 = k + 1 k = 7
4 + 7 + kn + 1 = 4 + 7 + 7.1 + 1 = 19
∗) 6 r = 9 = k + 1 5
4 + 7 + kn + 1 = 4 + 7 + 8.1. + 1 = 20
S x
1
1 x
2
≥ x
1
2 x
3
≥ x
2
3 x
n
≥ x
n−1
n v
r
v
r
r
v
r

=























n(r −1) + 1, r ≤ x
1
(n − 1)(r −1) + 1 + x
1
, x
1

< r ≤ x
2
(n − 2)(r −1) + 1 + x
1
+ x
2
, x
2
< r ≤ x
3

(1)(r −1) + 1 + x
1
+ x
2
+ + x
n−1
, x
n−1
< r ≤ x
n
x x [x]
x
m n
p + 1 p =

(m − 1)
n

.

p
np ≤ n

m − 1
n

= m−1 < m
26 m = 26 7
p =

25
7

= 3 4
X n
r ≤ n X
r X r = n
X
X = {A, A, B, B, B, C, C} AABCBBC
n
i
(i = 1, 2, , k) r n k + 2 n
1
+ n
2
+
+ n
k
= r ≤ n P (n; n
1

, n
2
, , n
k
) ≡
P (n, r)
n
1
!n
2
! n
k
!
P (n, r) =
P (n, n)
(n − r)!
P (n; n
1
, n
2
, , n
k
) = P (n; n
1
, n
2
, , n
k
, n − r)
P (18; 3, 4, 6) =

P (18, 3 + 4 + 6)
3!4!6!
=
P (18, 13)
3!4!6!
=
18!
3!4!6!5!
=
P (18; 3 + 4 + 6 + 5)
3!4!6!5!
= P (18; 3, 4, 6, 5)
X n
i
i (i = 1, 2, , k) P (n; n
1
, n
2
, , n
k
)
n = n
1
+ n
2
+ + n
k
p X n
X P (n, n) X
n

1
1 n − n
1
n
1
1 n −n
1
n
1
!
i (i = 2, 3, , k) q = n
1
!n
2
! n
k
!
p =
P (n, n)
q
= P (n; n
1
, n
2
, , n
k
)
X = {C, E, E, I, M, M, O, T, T }
X
P (9, 1, 2, 1, 2, 1, 2) =

9!
1!2!1!2!1!2!
= 45360
4
3 5 18
1
P (18; 4, 3, 5) =
18!
4!3!5!6!
= 514594080
X n S X
r S
r X r = n
r X n
1
1 n
2
2 n
k
k C(n; n
1
, n
2
, , n
k
)
n
1
+ n
2

+ + n
k
= r
C(n; n
1
, n
2
, , n
k
) = C(n, n
1
)C(n − n
1
, n
2
) C(n − n
1
− n
2
− − n
k−1
)
=
n!
n
1
!n
2
! n
k

!(n − r)!
=
P (n, r)
n
1
!n
2
! n
k
!
C(n; n
1
, n
2
, , n
k
) = P (n; n
1
, n
2
, , n
k
) n
1
+ n
2
+
+ n
k
= r ≤ n

17 5
5 4, 4, 2, 5 1
16 16 17
C(17; 4, 4, 2, 5, 1) =
17!
4!4!2!5!1!1!
n p
1
n
1
p
2
n
2
p
k
n
k
n
i
(i = 1, 2, , k)
k

i=1
p
i
n
i
= n
C(n;

p
1
  
n
1
, n
1
,
p
2
  
n
2
, n
2
, ,
p
k
  
n
k
, n
k
)
p
1
!p
2
! p
k

!
=
n!
[p
1
!(n
1
!)
p
1
][p
2
!(n
2
!)
p
2
] [p
k
!(n
k
!)
p
k
]
(i)
C(12; 4, 4, 4)
(ii)
C(12; 4, 4, 4)/3!
(ii)

C(12; 4, 4, 4)
3!
.4!
n r
r n
r n
r n n
r
n r r
n r
r
n C(n + r −1, r)
A n(A)
| A |
n(A ∪ B) = n(A) + n(B) − n(A ∩ B)
A B A∪B
n(A) n(B) n(A ∩ B)
A X A X A

A B X
n

(A ∪ B)


= n(X) − n(A ∪ B) = n(X) − [n(A) + n(B) + n(A ∩ B)]
(A ∪ B)

= A


∩ B

n(A

∩ B

) = n(X) − [n(A) + n(B)] + n(A ∩ B)
x X A
X x A 1 x A
0 x A
A
1
, A
2
, , A
m
X
n(A

1
∩ A

2
∩ ∩ A

m
) = n(X) − S
1
+ S
2

− + (−1)
m
S
m
S
k
k
m
(S
1
= n(A
1
) + n(A
2
) + + n(A
m
); S
2
=

i,j=1,m
i=j
n(A
i
∩ A
j
), )
x X
x
2

(i) x m
(ii) x r m r ≥ 1
A
1
, A
2
, , A
r
x 1
x 0
S
k
=

n(A
i
1
∩ A
i
2
∩ ∩ A
i
k
) (k = 1, 2, , m)
x
1 − C(r, 1) + C(r, 2) − C(r, 3) + + (−1)
r
C(r, r) = (1 − 1)
r
= 0

n(A
1
∪ A
2
∪ ∪ A
m
) = S
1
− S
2
+ + (−1)
m−1
S
m
n(A
1
∪A
2
∪ ∪A
m
) = n(X) −n(A

1
∩A

2
∩ ∩A

m
)

1
2
3
4
5
6
7
8
9
10
2.1.5 2.1.7 2.1.8
2.1.9 2.1.10
ABEUEBA
7 8
2
n

n + 1
2

4 0 9 9
2 8 3 7
4 (9).(9).(8).(7) = 4536
11 (1)
(1)
N 2k
k = 1 (1)
k ≥ 2
N = a
2k−1

.10
2k−1
+ a
2k−2
.10
2k−2
+ + a
k
.10
k
+ a
k
.10
k−1
+ + a
2k−2
.10
1
+ a
2k−1
.10
0
= a
2k−1
(10
2k−1
+ 10
0
) + (a
2k−2

.10
2k−2
+ + a
2k−2
.10
1
) =
a
2k−1
.P + Q
P = 100 001
  
2k
= 11. 9090 9091
  
2k−2
Q = a
2k−2
.10
2k−2
+ + a
2k−2
.10
1
Q 11 n 11
1
0 1
n
2.1.2


n + 1
2

−1 =

n − 1
2

1 0
2
[
n−1
2
]
100000
3 4
5
100000
0 00000
1 00001
2 00002
99999 99999
5 3
5 4 4
5 3
{0, 1, 2, 6, 7, 8, 9}
(5).(4).(3).(7).(7) = 2940
441000
n n 1 n
441000 = (2

3
).(3
2
).(5
3
).(7
2
).
(2
a
).(3
b
).(5
c
).(7
d
)
0 ≤ a ≤ 3; 0 ≤ b ≤ 2; 0 ≤ c ≤ 3; 0 ≤ d ≤ 2 a
4 b 3 c 4 d 3
(4).(3).(4).(3) − 2 = 142
N N
N = p
n
1
1
p
n
2
2
p

n
k
k
p
1
, p
2
, , p
k
2.1.6 N
(n
1
+ 1)(n
2
+ 1) (n
k
+ 1) − 2
n
i
i
i = 1, 2, , k
i p
i
p
i
N 2.1.7
N 1 2.1.7
(n
1
+ 1)(n

2
+ 1) (n
k
+ 1) − 1
441000 m n
m > 1, n > 1 m, n 1 m n
X = {2
3
; 3
2
; 5
3
; 7
2
}
441000 X
m n
2 m n
X 441000 m, n
X 2 m.n
n.m X
X = {2
3
} + {3
2
, 5
3
, 7
2
} = {3

2
} + {2
3
, 5
3
, 7
2
}
= {7
2
} + {2
3
, 3
2
, 5
3
}
= {2
3
, 3
2
} + {5
3
, 7
2
} = {2
3
, 5
3
} + {3

2
, 7
2
}
= {2
3
, 7
2
} + {3
2
, 5
3
}
4 + 3 = 7 = 2
4−1
− 1
2.1.9 N = p
n
1
1
p
n
2
2
p
n
k
k
p
1

, p
2
, , p
k
(k ≥ 2) N = m.n m, n
2
k−1
− 1 (m > 1, n > 1)
k
k = 2
k ≥ 3 k
Z = {a
1
, a
2
, , a
k−1
, a
k
} 2
k−1
−1
(k −1)
Z
Z = {a
k
} ∪ {a
1
, a
2

, , a
k−1
}
≡ {a
k
} ∪ W
W 2
k−2
− 1
a
k
Z
Z
1 + (2
k−2
− 1).2 = 2
k−1
− 1().
0
1 X n
n X Y = {0, 1}
n
X r = 2
n
2
r
f f
0 1
n = 6, f(101101) = f(010010) f
4 X = {00; 01; 10; 11}

Y = {0; 1}
a)f
1
(00) = f
1
(11) = f
1
(01) = f
1
(10) = 0
b)f
2
(00) = f
2
(11) = f
2
(01) = f
2
(10) = 1
c)f
3
(00) = f
3
(11) = f
3
(01) = f
3
(10) = 1
d)f
4

(00) = f
4
(11) = f
4
(01) = f
4
(10) = 0
n
2.1.12 X
r
2
= 2
n−1
(ς, ς

)
ς

ς 0 1
0 1 2
r
2
0
n 0 m
n + m
m
n
n + m m
m n + m
C(n + m, m)

m n
n A
1
, A
2
, , A
m
A
i
A
i
A
i
A
i
A
i
n ≤ 2
m
x m
a(x) = (x
1
, x
2
, , x
m
) x
i
= 1 x A
i

x
i
= 0
x A
i
f : S −→ T = {(x
1
, x
2
, , x
m
) | x
i
∈ {0, 1}}
x y f(x) f(y)
2
m
x
i
(x
1
, x
2
, , x
m
) 0 1
n ≤ 2
m
n(A) ≤ n(B) n(A) ≥ n(B)
n(A) = n(B)

f A B
x y A f(x), f(y)
B
A B A m B
n (n ≥ m).
P (n, m) P (n, m)
n
a)n = 14
b)n = 6
a) P (14, 10) 2.2.1
b)
P (10, 6)
n
n 1, 2, , n
n!
ABCD BCDA
(n −1)!
A
1
n
n − 1
n −1 (n −1)!
n p q
p q
p n − 1 q p
q n
n − 1
2
n
2

(n − 1)!
2
n n
(n −1)! n
n! n n!(n − 1)!
n 1
2 n
2
n −3 n −3 (n −3)!
2.(n − 3)!
r
n
r C(n, r)
(r − 1)! (n −r −1)!
C(n, r)(r − 1)!(n − r −1)!
m
n (m < n)
n n
(n −1)! n
n
m n P (n, m)
(n − 1)!P (n, m)
C(m + n, 2) − C(m, 2) − C(n, 2) = m.n
m n
m.n
C(m + n, 2)
C(m, 2) C(n, 2)
C(n, r) = C(n − 1, r) + C(n − 1, r −1)
X n Y