Solving Combinatorial Problems Using Binomial Coefficients

Bài 1. Các bài toán v công th c t h p, ch nh h p
237
CHƯƠNG III. T H P, XÁC SU T VÀ S PH C
BÀI 1. CÁC BÀI TOÁN V CÔNG TH C T H P, CH NH H P
I. D NG 1: CH NG MINH NG TH C
k
nC B NG O HÀM
1. Các bài t p m u minh h a:
Bài 1. Ch ng minh r ng: −1 2 n n 1
n n nC + 2C + ...+ n.C = n2
Gi i
Xét: (1 + x)n
= o 1 2 2 3 3 n 1 n 1 n n
n n n n n nC C x C x C x ... C x C x− −
+ + ⋅ + ⋅ + + +
L y o hàm c 2 v ta có: ( )n 1 1 2 3 2 n n 1
n n n nn 1 x C 2C x 3C x nC x
− −
+ = + ⋅ + ⋅ + + ⋅…
Th x = 1 vào ng th c trên ta có: 1 2 n n 1
n n nC 2C ... n.C n2 −
+ + + =
Bài 2. Ch ng minh r ng: −
− −2 3 n n 2
n n n2.1.C + 3.2.C + ...+ n(n 1)C = n(n 1)2
Gi i
Xét: ( )1
n
x+ = o 1 2 2 3 3 n 1 n 1 n n
+ + ⋅ + ⋅ + + +
L y o hàm c 2 v ta có: ( )n 1 1 2 3 2 n n 1
n n n nn 1 x C 2C x 3C x nC x
− −
+ = + ⋅ + ⋅ + + ⋅…
L i l y o hàm ta có: ( )( )n 2 2 3 n n 2
n n nn n 1 1 x 2C 3.2.C .x n(n 1)C .x
− −
− + = + + + −…
Th x = 1 vào ng th c trên ta có: 2 3 n n 2
n n n2.1.C 3.2.C ... n(n 1)C n(n 1)2 −
+ + + − = −
Bài 3. ( thi TS H kh i A −−−− 2005): Gi i phương trình:
( )− −1 2 2 3 3 4 2n 2n+1
2n+1 2n+1 2n+1 2n+1 2n+1C 2.2C + 3.2 C 4.2 C + ...+ 2n + 1 2 C = 2005
Gi i
Xét ( )2 1 0 1 2 2 2 1 2 1
2 1 2 1 2 1 2 1 2 11 ... ...
n k k n n
n n n n nx C C x C x C x C x
+ + +
+ + + + ++ = + + + + + +
L y o hàm c 2 v ta có:
( )( ) ( )2 1 2 1 2 1 2
2 1 2 1 2 1 2 12 1 1 2 ... ... 2 1
n k k n n
n n n nn x C C x kC x n C x− +
+ + + ++ + = + + + + + +
Thay x = −2 vào ng th c ta có:
( ) ( ) ( ) ( )1 21 2 2 1
2 1 2 1 2 1 2 12 1 2.2 ... 2 ... 2 2 1
k nk n
n n n nn C C kC n C
− +
+ + + ++ = − + + − + + − +
Phương trình ã cho ⇔ 2n + 1 = 2005 ⇔ n = 1002

Chương III. T h p, Xác su t và S ph c −−−− Tr n Phương
238
Bài 4. Gi i phương trình:
( ) ( ) ( )−
− − − − −
k2 3 k 2 k 2n-1 2n+1
2n+1 2n+1 2n+1 2n+12C 3.2C + ...+ 1 k k 1 2 C + ... 2n 2n + 1 2 C = 110
Gi i
Xét ( ) ( )2 1 0 1 2 2 2 1 2 1
2 1 2 1 2 1 2 1 2 11 ... 1 ...
n k k k n n
n n n n nx C C x C x C x C x
+ + +
+ + + + +− = − + − + − + −
L y o hàm c 2 v ta có:
( )( ) ( ) ( )2 1 2 1 2 1 2
2 1 2 1 2 1 2 12 1 1 2 ... 1 ... 2 1
n k k k n n
n n n nn x C C x kC x n C x− +
+ + + +− + − = − + − + − + − +
L i l y o hàm c 2 v ta có: ( )( )2 1
2 2 1 1
n
n n x
−
+ − =
( ) ( ) ( )2 3 2 2 1 2 1
2 1 2 1 2 1 2 12 3 ... 1 1 ... 2 2 1
k k k n n
n n n nC C x k k C x n n C x− + −
+ + + += − + + − − + − +
Thay x = 2 vào ng th c ta có: ( )2 2 1n n− + =
( ) ( ) ( )2 3 2 2 1 2 1
2 1 2 1 2 1 2 12 3.2 ... 1 1 2 ... 2 2 1 2
k k k n n
n n n nC C k k C n n C− − +
+ + + += − + + − − + − +
Phương trình ã cho ⇔ ( ) 2
2 2 1 110 2 55 0 5n n n n n+ = ⇔ + − = ⇔ =
2. Các bài t p dành cho b n c t gi i:
Bài 1. Ch ng minh r ng: 0 1 2
3 5 ... (2 1) ( 1)2n n
n n n nC C C n C n+ + + + + = +
Bài 2. Ch ng minh r ng: 1 1 2 2 3 3 1
2 2.2 3.2 ... . .3n n n n n
n n n nC C C n C n− − − −
+ + + + =
Bài 3. Ch ng minh r ng: 1 2 3 4 1
2 3 4 ... ( 1) . 0n n
n n n n nC C C C n C−
− + − + + − =
Bài 4. Ch ng minh r ng:
( ) ( ) ( )1 0 2 1 3 2 1 1 1 2 1
4 1 4 2 4 ... 1 2.2 .. .2
nn n n n n n
n n n n n n nn C n C n C C C C n C− − − − −
− − + − − + − = + + +
Bài 5. Ch ng minh r ng: ( ) ( ) ( )
( )
( )[ ]
2 2 21 2
2
2 1 !
2
1 !
n
n n n
n
C C n C
n
−
+ +…+ =
−
∀n ≥ 2
( ) ( )
( )
( )
2 3
2 3
2 1
1
1 1 1
n
n n n
n
C C n C
n n n
−
+ + + =
− − −
… ∀n ≥ 2
Bài 7. Ch ng minh r ng: ( ) 11
1
tg 1 tg
n
nk k
n
k
kC x n x
−−
=
= +∑ ∀n ≥ 2
Bài 8. Ch ng minh r ng: ( )1 2 2 2 3 2 2
2 3 ... 1 2n n
n n n nC C C n C n n −
+ + + + = +
Bài 9. Ch ng minh r ng: ( ) ( ) ( ) 11 2 1
1 2 ... 1 0
nn n n
n n n nnC n C n C C
−− −
− − + − − + − =
Bài 10. CMR: ( ) ( ) ( )1 20 1 1 2 1 1
1 2 1 2 .2 ... 1 2 ... 2
n n n k k k n n
n n n nC C kC nC n
− − − − −
− + − − + − + + =

239
II. D NG 2: CH NG MINH NG TH C
k
nC B NG TÍCH PHÂN
1. Các bài t p m u minh h a:
−n+1
1 2 n
n n n
2 11 1 1
1 + C + C + ...+ C =
2 3 n + 1 n + 1
Gi i
Xét (1 + x)n
= o 1 2 2 3 3 n 1 n 1 n n
+ + ⋅ + ⋅ + + +
Ta có: ( )
( )n 11 n 11
n
0
0
1 x 2 1
1 x dx
n 1 n 1
+ ++ −
+ = =
+ +∫
M t khác: ( )
1
o 1 2 2 3 3 n 1 n 1 n n
n n n n n n
0
C C x C x C x ... C x C x dx− −
+ + ⋅ + ⋅ + + + =∫
−
−
n+1
1 2 n
n n n
( 1)1 1 n
C C + ...+ C =
2 3 n + 1 n + 1
Gi i
Ta có : (1 − x)n
= 0 1 2 2 n n n
n n n nC C x C x ... ( 1) C x− + + + −
⇒
2 2 n 1
n 0 1 n n n 1 2 n
n n n n n n
0 0
( 1)1 1
(1 x) dx C C x ... ( 1) C x dx C C ... C
2 3 n 1
+
−
 − = − + + − = − + +  +∫ ∫
M t khác
12 n 1
n
0 0
(1 x) 1
(1 x) dx
n 1 n 1
=
−
− = =
+ +∫ ⇒ ( pcm)
( )
−n+1
0 1 2 n
n n n n
1 1 1 1 2 1
C + C + C + …+ C =
3 6 3 3n + 3 3 n + 1
Gi i
Xét P(x) = ( ) ( )
n
2 3 2 0 1 3 2 6 n 3n
n n n nx 1 x x C C x C x C x+ = + ⋅ + ⋅ + + ⋅…
Ta có:
( ) ( ) ( )
1 1 1
n n
2 3 3 3
0 0 0
1
P(x)dx x 1 x dx 1 x d 1 x
3
= + = + +∫ ∫ ∫
( )
( )
n 1
3 n 1
1 1 x 2 1
3 n 1 3 n 1
+
+
+ −
= =
+ +

240
M t khác: ( )
1 1
0 2 1 5 n 3n 2
n n n
0 0
P(x) dx C x C x C x dx+
= ⋅ + ⋅ + + ⋅∫ ∫ … =
=
1
0 3 1 6 n 3n 3
n n n
0
C x C x C x
3 6 3n 3
+
 ⋅ ⋅ ⋅
+ + + 
+  
… 0 1 2 n
n n n n
1 1 1 1
C C C C
3 6 3 3n 3
= + + + +
+
…
V y
( )
n 1
0 1 2 n
n n n n
1 1 1 1 2 1
C C C C
3 6 3 3n 3 3 n 1
+
−
+ + + + =
+ +
…
2. Các bài t p dành cho b n c t gi i:
n
1 2 n n
n n
C1 1 1
1 C C ... ( 1)
2 3 n 1 n 1
− + − + − =
+ +
n
0 1 2 n
n n n n
( 1)1 1 1 1
C C C ... C
2 4 6 n 2 2(n 1)
−
− + − + =
+ +
n n
0 2 1 3 2 n 1 n
n n n n
( 1) 1 ( 1)1 1
2C 2 C 2 C ... 2 C
2 3 n 1 n 1
+− + −
− ⋅ + ⋅ − + ⋅ =
+ +
n 1
0 2 1 3 2 n 1 n
n n n n
3 11 1 1
2C 2 C 2 C ... 2 C
2 3 n 1 n 1
+
+ −
+ ⋅ + ⋅ + + ⋅ =
+ +
Bài 5. Ch ng minh r ng: ( ) ( )
( )
0 1 2 3 2 !!1 1 1 1... 1
3 5 7 2 1 2 1 !!
n n
n n n n n
n
C C C C C
n n
− + − + + − =
+ +
( )n 1 n nn 1
k k k 1
n n
k 0 k 0
1 e 1 2 1
C C e
n 1 k 1 n 1 k 1
+ +
+
= =
+
+ = +
+ + + +
∑ ∑
( )0 1 2 11 1 1...
2 3 1 1
n
n
n n n nC C C C
n n
−
− + − + =
+ +
Bài 8. Ch ng minh r ng: ( )1 2 3
3 7 ... 2 1 3 2n n n n
n n n nC C C C+ + + + − = −
( ) ( )
k kn n 2n 2 n 1
n n
k 1 n 1
k 0 k 0
C C 2 3
k 1 k 1 2 n 1 2
+ +
+ +
= =
−
− =
+ + +
∑ ∑
Bài 10. t Sn =
1 1 1
1
2 3 n
+ + + +… . Ch ng minh r ng:
( )
( )n 1
n 11 2 n 1
n n n 1 n n 2 n 1
1
S C S C S 1 C S
n
−
− −
− −
−
− + − + − =…
Bài 11. Ch ng minh: ( )n 11 2 3 n
n n n n
1 1 1 1 1 1
C C C 1 C 1
1 2 3 n 2 n
−
⋅ − ⋅ + ⋅ − + − ⋅ ⋅ = + + +… …

241
III. D NG 3: CH NG MINH NG TH C
k
nC B NG NH NGHĨA
1
1
k k
n n
nC C
k
−
−= ( k n< ) ; ( ) 1
1m m
n m n mnC m C +
+ += + ; m k k m k
n m n n kC C C C −
−⋅ = ⋅ (k ≤ m ≤ n) ;
( )
2 3
1
1 2 1 1
1
2 3 ... ...
2
p n
n n n n
n p n
n n n n
C C C C n n
C p n
C C C C− −
+
+ + + + + + = ;
( ) ( )
1
11 2 3
1
0
2 3 ... 1 1
n
n kn k
n n n n n
k
C C C nC n C
−
−
−
=
− + − + − = −∑ ; 1 1
2 2 2 2
1
2
n n n
n n nC C C− +
++ = ;
0 1 2
1 2 3 1 1
2 3 4 2 2 2
1
... ...
2
k n
n n n n n
k n
n n n n k n
C C C C C
C C C C C+ +
+ + + + + +
+ + + + + + = ; 1 1 1
22m m m m
n n n nC C C C+ − +
++ + = ;
IV. D NG 4: CH NG MINH B NG CÔNG TH C
1
1 1;− −
− −= + =k n k k k k
n n n n nC C C C C
1
1 2 1 1...k k k k k k
n n n k k nC C C C C C +
− − + ++ + + + + = ; 1 2 3
33 3k k k k k
n n n n nC C C C C− − −
++ + + =
1 2 3 2 3
2 32 5 4k k k k k k
n n n n n nC C C C C C+ + + + +
+ ++ + + = + ; 1
0
m
k m
n k n m
k
C C+ + +
=
=∑
1 2 3 4
44 6 4k k k k k k
n n n n n nC C C C C C− − − −
++ + + + =
V. D NG 5: CH NG MINH B NG KHAI TRI N NEWTON
0 1
... 2n n
n n nC C C+ + + = ; 1 3 2 1 0 2 2 2 1
2 2 2 2 2 2... ... 2n n n
n n n n n nC C C C C C− −
+ + + = + + + =
0 1 1 1
3 3 ... 3 4n n n n n
n n n nC C C C− −
+ + + + = ; 0 1 2 2 3 3
6 6 6 ... 6 7n n n
n n n n nC C C C C+ + + + + =
( )0 1 2 3
... 1 0
n n
n n n n nC C C C C− + − + + − = ; 0 1 1 2 2 3 1 1
2 2 .2 2 .3 ... 2 . .3n n n
n n n nC C C nC n− −
+ + + + =
0 2 2 1 3 2 1
... ... ... ...k k
n n n n n nC C C C C C +
+ + + + = + + + +
0 1 2 2 3 3 2 1 2 1 2 2
2 2 2 2 2 210 10 10 ... 10 10 81n n n n n
n n n n n nC C C C C C− −
− + − + − + =
( ) ( )0 1 1 2 2
2 2 2 ... 1 2 ... 1 1
k nn n n n k k n
n n n n nC C C C C− − −
− + − + − + + − =
( )0 1 1 2 2 0 1 2 2
4 4 4 ... 1 2 2 .. 2
nn n n n n n
n n n n n n n nC C C C C C C C− −
− + − + − = + + + +
0 2 1 3 2 1 12 2 2 2 3 1...
1 2 3 1 1
n n n
n n n nC C C C
n n
+ +
−+ + + + =
+ +
( )0 2 2 4 4 2 2 2 1 2
2 2 2 23 3 ... 3 2 2 1n n n n
n n n nC C C C −
+ + + + = +
( )0 2 2 4 4 2000 2000 2000 2001
2001 2001 2001 20013 3 ... 3 2 2 1C C C C+ + + + = −

242
VI. D NG 6: CH NG MINH NG TH C B NG CÁCH NG NH T H S THEO 2
CÁCH KHAI TRI N
0 1 1 1 1 0
. . ... . .k k k k k
n m n m n m n m m nC C C C C C C C C− −
++ + + + =
0 1 1
2. . ... .k k n k n n k
n n n n n n nC C C C C C C+ − +
+ + + =
( ) ( ) ( )
2 2 20 1
2... n n
n n n nC C C C+ + + =
( ) ( ) ( ) ( )
2 2 2 20 1 2 2 1
2 1 2 1 2 1 2 1... 0n
n n n nC C C C +
+ + + +− + − − =
( ) ( ) ( ) ( ) ( )
2 2 2 20 1 2 2
2 2 2 2 2... 1
nn n
n n n n nC C C C C− + − + = −
VII. D NG 7: PHƯƠNG TRÌNH, H PHƯƠNG TRÌNH, BPT CH A ; ;k k
n n nA C P
1. Gi i các phương trình sau ây:
3 2
2 20n nC C= ;
4
3 4
1
24
23
n
n
n n
A
A C −
+
=
−
; 3
5
5
720n
n n
P
A P
+
−
= ; 1 3
172 72n nA A +− = ;
1 2 3 2
6 6 9 14n n nC C C n n+ + = − ; ( )2 2
72 6 2n n n nP A A P+ = + ;
5 6 7
5 2 14
n n n
C C C
− = ;
4 3 2
1 1 2
5 0
4n n nC C A− − −− − = ; 1 1
1 1 1: : 5:5:3m m m
n n nC C C+ −
+ + + = ; 3 2
14n
n nA C n−
+ = ;
3 3
8 65n
n nC A+
+ += ; 1
2 2 235 132n n
n nC C−
−= ;
4 5 6
1 1 1
n n n
C C C
− = ;
( )2 4 4 2
1 1 4 1n n
n n nn C xC x C− −
− −+ = + ; 1 1
2 2 13 2n n
n nC C− −
+= ; 3 2 1
14 n
n n nA C C −
+ =
2. Gi i các b t phương trình sau ây:
( )
4
4 42
2 !
n
n
A
Pn
+
≤
+
;
( )
4
4 143
42 !
n
n
A
Pn
+
≤
+
;
2
1
2
1
2
n
n
n
n
A
P
C
−
+
−
≥ ;
3
3
1
195 0
4
n
n n
A
P P
+
+
− > ;
4
4
2
143 0
4
n
n n
A
P P
+
+
− > ;
2 2 3
2
61 10
2 n n nA A C
x
− ≤ + ; 11
13 13
n m
C C −
≥ ; 4 3 2
1 1 2
5 0
4n n nC C A− − −− − = ; 1 1
112 162n nC C−
++ ≥ ;
1 3
172 72n nA A +− ≤ ; 2 2
12 3 30n nC A+ + < ; 3 1
1 1100 n
n nC C −
+ +≥ +
3. Gi i các h phương trình sau ây:
2 5 90
5 2 80
y y
x x
y y
x x
A C
A C
 + =

 − =
;
( ) ( )
( )
221 1 1 1
31 1
2 3
2 1
x y x y
x y x y
x y
x y
C C A C
C A
− − − −
− −
 + =


 = +
;
( ) ( )( )
1 1
1 1
3
2
3 2
1 1
6 14
32 1
x y
x y x y
x y
y
yx
x
C A
C A
C
x C
y y
− −
− −
−
−
 − = −




= + +
− −

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