Concrete Mathematics 6.1 Stirling Number Identities

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#算法

文章探讨了组合公式(6.15)和(6.16)，涉及子集大小的计算以及元素分配到非空子集的问题。公式(6.15)描述了从n个元素中选择k个与特定元素一起形成子集的组合数，然后将剩余的k个元素分配到m个非空子集中。而公式(6.16)涉及到将n个元素先划分为k个循环，再从中选取m个循环与其他元素组成新循环的方式。文章通过实例解释了这两个公式的差异及证明思路。

${n+1m+1}=∑k(nk){km}(6.15)\begin{Bmatrix}n+1 \\m+1\end{Bmatrix}=\sum_k \binom{n}{k}\begin{Bmatrix}k \\m\end{Bmatrix} \tag{6.15}$

We enumerate the size of the subset which $n + 1$ belongs to, which equals $n - k + 1$ . We first choose $n - k$ elements that is in the same subset with $n + 1$ , which is $(nk)\dbinom{n}{k}$ , then allocate the rest $k$ elements into $m$ non-empty subsets.

$[n+1m+1]=∑k[nk](km)(6.16)\begin{bmatrix}n+1 \\m+1\end{bmatrix}=\sum_k \begin{bmatrix}n \\k\end{bmatrix}\binom{k}{m} \tag{6.16}$

This looks almost the same as the previous one, but it’s much harder to consider the combinatorial explanation.
What the formula do is obvious. First divide the first $n$ elements into $k$ cycles, then choose $m$ cycles from it, and combine the rest $k - m$ cycles with $n + 1$ to form a new cycle. But the question is how to proof this approach is not heavy and not leaky.
Consider that we first divide the elements into $m + 1$ subsets $,Am+1A_1,A_2,\cdots,A_{m+1}$ . We assume that $n+1∈A1n+1\in A_1$ . Say $C(A_i)$ is a cycle formed with elements in $A_i$ . Take any specific $,C(Am+1)C(A_2),C(A_3),\cdots,C(A_{m+1})$ ,there are $A_1|-1)!$ ways to choose $C(A_1)$ . In our formula, with $,C(Am+1)C(A_2),C(A_3),\cdots,C(A_{m+1})$ fixed(after we times $k$ choose $m$ ), the subset $A_1$ is constitute of some of the cycles of its elements, which sum up to exactly $∑k[∣A1∣−1k]\sum_k\begin{bmatrix}|A_1|-1 \\k\end{bmatrix}$ times, which equals $A_1|-1)!$ .