MathcotさんのHomepageです



二項定理・組合せnCr

Author: Mathcot

初版: 2007.8.13

Last Update: 2008.04.05



順列 nPr

nPr=n!/r!=n(n-1)…(r+1) 
nPn=1
nP0=nP1=n!
nP(n-1)=n
n
nPr , r=0〜n
0
1
2
3
4
5
6
7
8
0P0=1
1P0=1P1=1
2P0=2P1=2, 2P2=1
3P0=3P1=6, 3P2=3, 3P3=1
4P0=4P1=24, 4P2=12, 4P3=4, 4P4=1
5P0=5P1=120, 5P2=60, 5P3=20, 5P4=5, 5P5=1
6P0=6P1=720, 6P2=360, 6P3=120, 6P4=30, 6P5=6, 6P6=1
7P0=7P1=5040, 7P2=2520, 7P3=840, 7P4=210, 7P5=42, 7P6=7, 7P7=1
8P0=8P1=40320, 8P2=20160, 8P3=6720, 8P4=1680, 8P5=336, 8P6=56, 8P7=8, 8P8=1

組合せ(Combination) nCr


nCr=n!/{(n-r)!r!} 
nCr=(n-1)Cr+(n-1)C(r-1)
nC0=nCn=1
nC1=nC(n-1)=n
n
nCr, r=0, n
1
2
3
4
5
6
7
8
9
10
1,1
1,2,1
1,3,3,1
1,4,6,4,1
1,5,10,10,5,1
1,6,15,20,15,6,1
1,7,21,35,35,21,7,1
1,8,28,56,70,56,28,8,1
1,9,36,84,126,126,84,36,9,1
1,10,45,120,210,252,210,120,45,10,1
11
12
13
14
15
16
17
18
19
20
1,11,55,165,330,462,462,330,165,55,11,1
1,12,66,220,495,792,924,792,495,220,66,12,1
1,13,78,286,715,1287,1716,1716,1287,715,286,78,13,1
1,14,91,364,1001,2002,3003,3432,3003,2002,1001,364,91,14,1
1,15,105,455,1365,3003,5005,6435,6435,5005,3003,1365,455,105,15,1
1,16,120,560,1820,4368,8008,11440,12870,11440,8008,4368,1820,560,120,16,1
1,17,136,680,2380,6188,12376,19448,24310,24310,19448,12376,6188,2380,680,136,17,1
1,18,153,816,3060,8568,18564,31824,43758,48620,43758,31824,18564,8568,3060,816,153,18,1
1,19,171,969,3876,11628,27132,50388,75582,92378,92378,75582,50388,27132,11628,3876,969,171,19,1
1,20,190,1140,4845,15504,38760,77520,125970,167960,184756,167960,125970,77520,38760,15504,4845,1140,190,20,1
21
1,21,210,1330,5985,20349,54264,116280,203490,293930,352716,352716,293930,203490,116280,54264,20349,5985,1330,210,21,1
22
1,22,231,1540,7315,26334,74613,170544,319770,497420,646646,705432,
646646,497420,319770,170544,74613,26334,7315,1540,231,22,1
23
1,23,253,1771,8855,33649,100947,245157,490314,817190,1144066,1352078,
1352078,1144066,817190,490314,245157,100947,33649,8855,1771,253,23,1
24
1,24,276,2024,10626,42504,134596,346104,735471,1307504,1961256,2496144,,2704156,
2496144,1961256,1307504,735471,346104,134596,42504,10626,2024,276,24,1

二項係数の計算
[Maple10]
n:=10:for r from 0 by 1 to n do (n,r),n!/r!*(n-r)! end do;


10,0,1
10,1,10
10,2,45
10,3,120
10,4,210
10,5,252
10,6,210
10,7,120
10,8,45
10,9,10
10,10,1
[フリーソフト Risa/Asir ]
[Program] def comb(A,B){for(I=1, C=1; I<=B; I++) C *= (A-I+1)/I; return C;}$
[実行] comb(1000,300);

50C5=2118760, 50C10=10272278170, 50C20=47129212243960,
50C25=126410606437752
100C10=17310309456440,
100C20=535983370403809682970,
100C50=100891344545564193334812497256
200C100=
90548514656103281165404177077484163874504589675413336841320

400C200=
1029525001354144329729758803204019867572109253810776482348490595759233323726519585983365955189764929
51564048597506774120

500C250=
1167443157882776829209347347621766196592300811803114461241002849578111126736084737156664177755216053
76810865902709989580160037468226393900042796872256

1000C500=
2702882409454365695156146936259752754961520084465482870073928751066254287055221938986124839245023701
6536260608502154610480220975005067991754989421969951847542366548426375173335616246407973788734436457
4161119497604571044985756287880514600994219426752366915856603136862602484428109296905863799821216320



二項定理
(1+x)n=Σ[k=0,n] nCk xk …(A)

(A)でx=1とおけば
2n=
Σ[k=0,n] nCk = nC0+nC1+nC2+nC3+ … +nC(n-1)+nCn

nC0+nC1+nC2+nC3+ … +nC(n-1)+nCn=2n

nC0=nCn=1であるから
nC1+nC2+nC3+ … +nC(n-1) = 2n -2
nC1+nC2+nC3+ … +nC(n-1)+nCn = 2n -1

(A)をx=-1とおけば
0=Σ[k=0,n] (-1)k nCk
nC0-nC1+nC2-nC3+ … +(-1)n nCn = 0

(A)でxで微分
n(1+x)(n-1) = Σ[k=1,n] nCk kx(k-1) …(A)
x=1,x=-1を代入して
n 2(n-1)=Σ[k=1,n] k(nCk) = nC1+2(nC2)+3(nC3)+ … +n nCn
0=Σ[k=1,n] k{(-1)^k}(nCk) = nC1-2(nC2)+3(nC3)+ … + n(-1)(n-1) nCn

以上整理すると
公式

Σ[k=0,n] nCk=2^n
Σ[k=1,n-1] nCk=2^n-2
Σ[k=1,n] nCk=2^n-1
nC0+nC1+nC2+nC3+…+nC(n-1)+nCn=2^n
nC1+nC2+nC3+…+nC(n-1)=2^n -2
nC1+nC2+nC3+…+nC(n-1)+nCn=2^n -1
Σ[k=0,n] {(-1)^k}nCk=0
Σ[k=1,n] {(-1)^(k-1)}nCk=0
nC0-nC1+nC2-nC3+…+{(-1)^n}nCn=0
nC1-nC2+nC3+…+{(-1)^(n-1)}nCn=1
n=enen
n=odd
nC0+nC2+nC4+…+nC(n-1)+nCn=2^(n-1)
nC0+nC2+nC4+…+nC(n-1)=2^(n-1)
n=enen
n=odd
nC1+nC3+nC5+…+nC(n-1)=2^(n-1)
nC1+nC3+nC5+…+nC(n-1)+nCn=2^(n-1)
Σ[k=1,n] k(nCk)=n2^(n-1)
Σ[k=1,n] k{(-1)^k}(nCk)=n2^(n-1)
nC1+2(nC2)+3(nC3)+…+n(nCn)=n2^(n-1)
nC1-2(nC2)+3(nC3)+…+n{(-1)^(n-1)}(nCn)=0
n=odd
n=even
nC1+3(nC3)+…+n(nCn)=n2^(n-2)
nC1+3(nC3)+…+n{nC(n-1)}=n2^(n-2)
n=odd
n=even
2(nC2)+4(nC4)+…+(n-1){nC(n-1)} = n 2(n-2)
2(nC2)+4(nC4)+…+n(nCn) = n 2(n-2)

参考URL
[1]
[2]
[3]

 
inserted by FC2 system