A 1 A_1 A1事件 n 1 n_1 n1种方法, A 2 A_2 A2事件 n 2 n_2 n2种方法。
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就相当于在大小为n得序列中随机选一个元素选择r次
n个元素无放回得取r个元素排列为有重复的排列–>种类有 A n r A_n^r Anr种 A n r = n ( n − 1 ) ( n − 2 ) . . . ( n − r + 1 ) = n ! ( n − r ) ! A_n^r =n(n-1)(n-2)...(n-r+1)=\frac{n!}{(n-r)!} Anr=n(n−1)(n−2)...(n−r+1)=(n−r)!n! 特指:n==r,此类排列叫全排列
无放回取r次,第一次有n种选择,接下来有n-1次,以此类推,到第r此选择得时候还有n-r+1个元素,所以就有n-r+1个可能性
P n = n ( n − 1 ) × ( n − 2 ) × . . . × 3 × 2 × 1 = n ! P_n=n(n-1)\times(n-2)\times...\times 3\times 2\times 1=n! Pn=n(n−1)×(n−2)×...×3×2×1=n!
n个不同元素抽出r个不考虑顺序成为组合,总数为: C n r = A n r r ! = n ( n − 1 ) ( n − 2 ) . . . ( n − r + 1 ) r ! = n ! ( n − r ) ! r ! C_n^r=\frac{A_n^r}{r!}=\frac{n(n-1)(n-2)...(n-r+1)}{r!}=\frac{n!}{(n-r)!r!} Cnr=r!Anr=r!n(n−1)(n−2)...(n−r+1)=(n−r)!r!n!
其实这就是常见的二项式系数,例如 ( a + b ) n = ∑ r = 0 n C n r a r b n − r (a+b)^n=\sum_{r=0}^nC_n^ra^rb^{n-r} (a+b)n=∑r=0nCnrarbn−r
3.1下一步得拓展: r 1 + r 2 + . . . + r k = n r_1+r_2+...+r_k=n r1+r2+...+rk=n,把n个元素分成k各部使得 n i = r i ( i ∈ [ 1 , k ] ) n_i=r_i(i\in[1,k]) ni=ri(i∈[1,k]),结果为 n ! r 1 ! r 2 ! . . . r k ! \frac{n!}{r_1!r_2!...r_k!} r1!r2!...rk!n!
计算方法: 1.分成二元时结果为 C n r C_n^r Cnr,将二元元素看作 r 1 r_1 r1和 r t e a g e t 1 ( r 1 + r t e a g e t 1 = n ) r_{teaget_1}\quad (r_1+r_{teaget_1}=n) rteaget1(r1+rteaget1=n),将组合公式看成 C n r 1 C_{n}^{r1} Cnr1或 C n r t e a g e t 1 C_{n}^{r_{teaget_1}} Cnrteaget1 2.第二部分 r t e a g e t 1 r_{teaget_1} rteaget1继续分成二元 r 2 r_2 r2和 r t e a g e t 2 ( r 2 + r t e a g e t 2 = r t e a g e t 1 ) r_{teaget_2}\quad (r_2+r_{teaget_2}=r_{teaget_1}) rteaget2(r2+rteaget2=rteaget1)成为 C t e a g e t 1 r 2 C_{teaget_1}^{r_2} Cteaget1r2或 C t e a g e t 1 t e a g e t 2 C_{teaget_1}^{{teaget_2}} Cteaget1teaget2,这时可以看成把n个元素分成三部分 C n r 1 C t e a g e t 1 r 2 = C n r 1 C n − r 1 r 2 C_{n}^{r1}C_{teaget_1}^{r_2}=C_{n}^{r_1}C_{n - r_1}^{r_2} Cnr1Cteaget1r2=Cnr1Cn−r1r2,化简后得到: n ! ( n − r 1 ) ! r 1 ! × ( n − r 1 ) ! ( n − r 1 − r 2 ) ! r 2 ! = n ! r 1 ! r 2 ! r t e a g e t 2 ! \frac{n!}{(n-r_1)!r_1!}\times\frac{(n-r_1)!}{(n-r_1-r_2)!r_2!}=\frac{n!}{r_1!r_2!r_{teaget_2}!} (n−r1)!r1!n!×(n−r1−r2)!r2!(n−r1)!=r1!r2!rteaget2!n!,把 r t e a g e t 2 r_{teaget_2} rteaget2看成 r 3 r_3 r3可将最终三元答案记为 n ! r 1 ! r 2 ! r 3 ! \frac{n!}{r_1!r_2!r_3!} r1!r2!r3!n! 3.不严格的数学归纳一下,答案就出来了…(这里公式难打所以我们先这样做,规范化证明以后放出)r 1 + r 2 + . . . + r k = r r_1+r_2+...+r_k=r r1+r2+...+rk=r,把n个元素分成k各部使得 ∑ i = 1 k n i = n \sum_{i=1}^k n_i=n ∑i=1kni=n,需要取得r个元素,且要从 n i n_i ni中取得 r i r_i ri个,
简化思想,分开拿 直接每块拿一个照写 答案为: ∏ i = 1 k C n i r i \prod^{k}_{i=1} C_{n_i}^{r_i} ∏i=1kCniri
n个元素有重复的取r个,答案为: C n + r − 1 r C_{n+r-1}^r Cn+r−1r
C n 1 C_n^1 Cn1看做有重复的取一个 C n + 1 2 C_{n+1}^2 Cn+12就可以类推成有重复的拿两个,因为重复导致可以看作将n个中随机拿出一个复制一次然后放进去一次抽去两个 以此类推 最后答案就是上面的答案了
0 ! = 1 0!=1 0!=1 C n k = C n n − k C_n^k=C_n^{n-k} Cnk=Cnn−k ( 1 + x ) n = ∑ r = 0 n C n r x r (1+x)^n =\sum_{r=0}^nC^r_nx^r (1+x)n=∑r=0nCnrxr,上式x=1时, ∑ r = 0 n C n r = 2 n \sum_{r=0}^nC^r_n=2^n ∑r=0nCnr=2n C a 0 C b n + C a 1 C b n − 1 + . . + C a n C b 0 = C a + b n C_a^0C_b^n+C_a^1C_b^{n-1}+..+C_a^nC_b^0=C_{a+b}^n Ca0Cbn+Ca1Cbn−1+..+CanCb0=Ca+bn 证明挖坑 改日更新