组合数学做题笔记

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记号与约定

记号	意义
\(\mathbb{N}, \mathbf{N}\)	非负整数集
\(\mathbb{N}_+, \mathbf{N}_+\)	正整数集
\(\mathbb{Z}, \mathbf{Z}\)	整数集
\(\mathbb{Q}, \mathbf{Q}\)	有理数集
\(\mathbb{R}, \mathbf{R}\)	实数集
\(\mathbb{C}, \mathbf{C}\)	复数集
\(\sum\)	求和
\(\prod\)	求积

题目

Elementary Combinatorics （基本组合）

难度为 [1] / [1+]

1. [1] The number of subsets of an \(n\)-element set is \(2^n\).

这题的话，对每个元素单独讨论。每个元素有“选”与“不选”2种选择。总共有 \(n\) 个元素，乘法原理直接得到答案。

2. [1] A composition of \(n\) is a sequence \(\alpha = (\alpha_1, \alpha_2, \cdots, \alpha_k)\) of positive integers such that

\[\sum \alpha_i = n \]

The number of compositions of \(n\) is \(2^{n−1}\).

这题的话，用插板法。

在 \(n\) 个数中随机插 \(i\) 个板子，所以为

\[\sum_{i=0}^{n-1}\binom {n-1} {i}=2^{n-1} \]

6. [1] If \(S\) is an \(n\)-element set, then let \(\displaystyle\binom {S} {k}\) denote the set of all \(k\)-element subsets of S. Let \(\displaystyle\binom {n}{k} = \#\binom {S}{k}\), the number of
   k-subsets of an n-set. (Thus we are defining the binomial coefficient \(\displaystyle\binom {n}{k}\) combinatorially when \(n, k ∈ \mathbb N\).) Then

\[k !\binom n k=n(n-1) \cdots(n-k+1) \]

（第7题[1+]题意相近）

暴力就行。\(\displaystyle\binom n k = \dfrac{\mathrm{A}_n^k}{\mathrm A_k^k}\)，于是 \(k!\displaystyle\binom n k = \mathrm A_n^k = n(n-1) \cdots(n-k+1)\)。

8. [1] Let \(m, n ≥ 0\). How many lattice paths are there from \((0, 0)\) to \((m, n)\), if each step in the path is either \((1, 0)\) or \((0, 1)\)?

每一步向上 / 右 走一步，走到 \((m,n)\)，需要在 \((m+n)\) 次中选择 \(m\) 次走上。故有 \(\displaystyle\binom {n+m}m\) 种。

9. [1] For \(n > 0\), \(\displaystyle2\binom {2n-1} n = \binom {2n} n\).

这里的话，我们暴力。

\[2\binom {2n-1} n=2 \frac{\displaystyle \prod_{i=n}^{2n-1}i}{n!} \]

\[\begin{aligned} \binom {2n} n=&\frac{\displaystyle \prod_{i=n+1}^{2n}i}{n!},\\ =&\frac{\displaystyle \prod_{i=n}^{2n-1}i}{n!\cdot n\div n}\cdot 2,\\ =&\frac{\displaystyle \prod_{i=n}^{2n-1}i}{n!}\cdot 2,\\ =&2\binom {2n-1} n. \end{aligned} \]

10. [1+] For \(n ≥ 1\), \(\displaystyle\sum_{i=0}^{n}(-1)^i\binom n k = 0\).

暴力展开

\[\begin{aligned} \sum_{i=0}^{n}(-1)^i\binom n k=&\sum_{i=0,2\mid i}^{n}\binom n k-\sum_{i=0,2\not\mid i}^{n}\binom n k,\\ =&\cdots\\ =&0. \end{aligned} \]

（反正中间经过一次分类讨论就可以做出来）