Một số chuyên đề về tổ hợp dành
cho học sinh có năng khiếu toán
bậc trung học phổ thông class="bi x7 y9 w4 h9"
class="bi x7 y9 w4 h9"
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

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)

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

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

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

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

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 − 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

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

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


− + (−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

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
)

.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

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

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
} + {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

, 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

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

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

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