组合公式及推导
Pascal’s Triangle
$\begin{array}{c|cccccccccccccccccccccccccc}
\text{line}&\\\hline
0&&&&&&&&&&&&1&&&&&&&&&&\\\hline
1&&&&&&&&&&&1&&1&&&&&&&&&\\\hline
2&&&&&&&&&&1&&2&&1&&&&&&&&\\\hline
3&&&&&&&&&1&&3&&3&&1&&&&&&&\\\hline
4&&&&&&&&1&&4&&6&&4&&1&&&&&&\\\hline
5&&&&&&&1&&5&&10&&10&&5&&1&&&&&\\\hline
6&&&&&&1&&6&&15&&20&&15&&6&&1&&&&\\\hline
7&&&&&1&&7&&21&&35&&35&&21&&7&&1&&&\\\hline
8&&&&1&&8&&28&&56&&70&&56&&28&&8&&1&&\\\hline
9&&&1&&9&&36&&84&&126&&126&&84&&36&&9&&1&\\\hline
10&&1&&10&&45&&120&&210&&252&&210&&120&&45&&10&&1\\\hline
11&1&&11&&55&&165&&330&&462&&462&&330&&165&&55&&11&&1\end{array}$
其实，它的奇偶性就比较好玩：
$\begin{array}{c|cccccccccccccccccccccccccccccccc}
\text{line}&\\\hline
0&&&&&&&&&&&&&&&&1&&&&&&&&&&&&&&&\\\hline
1&&&&&&&&&&&&&&&1&&1&&&&&&&&&&&&&&\\\hline
2&&&&&&&&&&&&&&1&&0&&1&&&&&&&&&&&&&\\\hline
3&&&&&&&&&&&&&1&&1&&1&&1&&&&&&&&&&&&\\\hline
4&&&&&&&&&&&&1&&0&&0&&0&&1&&&&&&&&&&&\\\hline
5&&&&&&&&&&&1&&1&&0&&0&&1&&1&&&&&&&&&&\\\hline
6&&&&&&&&&&1&&0&&1&&0&&1&&0&&1&&&&&&&&&\\\hline
7&&&&&&&&&1&&1&&1&&1&&1&&1&&1&&1&&&&&&&&\\\hline
8&&&&&&&&1&&0&&0&&0&&0&&0&&0&&0&&1&&&&&&&\\\hline
9&&&&&&&1&&1&&0&&0&&0&&0&&0&&0&&1&&1&&&&&&\\\hline
10&&&&&&1&&0&&1&&0&&0&&0&&0&&0&&1&&0&&1&&&&&\\\hline
11&&&&&1&&1&&1&&1&&0&&0&&0&&0&&1&&1&&1&&1&&&&\\\hline
12&&&&1&&0&&0&&0&&1&&0&&0&&0&&1&&0&&0&&0&&1&&&\\\hline
13&&&1&&1&&0&&0&&1&&1&&0&&0&&1&&1&&0&&0&&1&&1&&\\\hline
14&&1&&0&&1&&0&&1&&0&&1&&0&&1&&0&&1&&0&&1&&0&&1&\\\hline
15&1&&1&&1&&1&&1&&1&&1&&1&&1&&1&&1&&1&&1&&1&&1&&1\\
\end{array}$
Hockey Stick Theorem
Theorem：$\dbinom{b+1}{a+1}=\dbinom{a}{a}+\dbinom{a+1}{a}+...+\dbinom{b}{a}$.
Proof:
$\binom{a}{a}+\binom{a+1}{a}=\binom{a+1}{a+1}+\binom{a+1}{a}=\binom{a+2}{a+1},$
$\binom{a}{a}+\binom{a+1}{a}+\binom{a+2}{a}=\binom{a+2}{a+1}+\binom{a+2}{a}=\binom{a+3}{a+1},$
$\binom{a}{a}+\binom{a+1}{a}+\binom{a+2}{a}+\binom{a+3}{a}=\binom{a+3}{a+1}+\binom{a+3}{a}=\binom{a+4}{a+1},$
and so on. Thus, induction can suffice.
引申：
\begin{aligned}
&1\times\binom{a+b}{b}+2\times\binom{a+b-1}{b}+\cdots+(a+1)\times\binom{b}{b}\\
=&\binom{a+b+1}{b+1}+\binom{a+b}{b+1}+\cdots+\binom{b+1}{b+1}=\binom{a+b+2}{b+2}
\end{aligned}
插板法 stars and bars / balls and urns
$n$ 个一样的东西，分成不同的 $m$ 组，运用插板法，有 $n-1$ 个可能位置和 $m-1$ 块板，有 $\dbinom{n-1}{m-1}$ 种方法
Vandermonde's Identity
Vandermonde's identity (or Vandermonde's convolution), named after Alexandre-Théophile Vandermonde, states that any combination of $k$ objects from a group of $(m+n)$ objects must have some $0\le r\le k$ objects from a group of $m$ objects and the remaining $(k-r)$ objects from a group of $n$.
$\dbinom{m+n}{k}=\sum_{r=0}^{k}\dbinom{m}{r}\dbinom{n}{k-r}$
圆环问题
A circle is divided into several sectors $S_{1}, S_{2}, \cdots, S_{n}$, and we use $k$ kinds of colours to paint these $n$ sectors $(n \geqslant 3, k \geqslant 3) .$ The painting rule is defined as follows:
(1) Every sector is painted in one colour:
(2) Adjacent sectors are never the same colour.
In how many ways can this circle be painted?
Let $a_{n}$ be the number of possible sets that meet all the conditions. Obviously, $a_{3}=k(k-1)(k-2)$.
For $n\ge4$, either $S_1=S_{n-1}$ or $S_1\ne S_{n-1}$.
If $S_1=S_{n-1}$, then there are $a_{n-2}$ possibilities for $S_1,S_2,...S_{n-1}$ and $k-1$ possibilities for $S_{n}$.
If $S_1\ne S_{n-1}$, then there are $a_{n-1}$ possibilities for $S_1,S_2,...S_{n-1}$ and $k-2$ possibilities for $S_{n}$.
Thus, $a_{n}=(k-2)a_{n-1}+(k-1)a_{n-2}$, so $a_{n}+a_{n-1}=(k-1)\left(a_{n-1}+(k-1)a_{n-2}\right)$.
Therefore, $a_{n}=(k-1)^{n}+(-1)^{n}(k-1)\quad(n\ge3)$.
好玩的性质
For $k\in\{0\}\cup\mathbb{Z}^+$, $\binom{k}{0}+\binom{k}{1}+...+\binom{k}{k}=2^{k}$.
If $m=2n$, $\begin{aligned}&\tbinom{m}{0}+\tbinom{m}{2}+...+\tbinom{m}{2n-2}+\tbinom{m}{2n}\\
=&\tbinom{m-1}{0}+\Big(\tbinom{m-1}{1}+\tbinom{m-1}{2}\Big)+...+\Big(\tbinom{m-1}{2n-3}+\tbinom{m-1}{2n-2}\Big)+1\\
=&\tbinom{m-1}{0}+\tbinom{m-1}{1}+...+\tbinom{m-1}{m-2}+\tbinom{m-1}{m-1}\\
=&2^{m-1}\end{aligned}$.
If $m=2n+1$, $\binom{m}{0}+\binom{m}{2}+...+\binom{m}{2n}=\binom{m}{1}+\binom{m}{3}+...+\binom{m}{2n+1}=2^{m-1}$.
