二项式系数 "$n$ 选取 $k$"：$$ \binom nk $$
对应的公式：$$ \binom nk = \frac {n!} {k! (n-k)!} = \frac {n^\underline k} {k!} $$
Stirling 数 "$n$ 轮换 $k$"/"$n$ 子集 $k$"：$$ {n \brack k}, {n \brace k} $$
$n$ 阶单位矩阵：$$ \mathbf{M} = \begin{bmatrix} 1 & 0 & 0 & 0 & \cdots & 0 & 0 \\ 0 & 1 & 0 & 0 & \cdots & 0 & 0 \\ 0 & 0 & 1 & 0 & \cdots & 0 & 0 \\ 0 & 0 & 0 & 1 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & 1 & 0 \\ 0 & 0 & 0 & 0 & \cdots & 0 & 1 \end{bmatrix} $$
下标测试：$a_{b\_c}, \mathrm{weight\_min}, \mathrm{weight\_max}$
连等式：\begin{align*} & D(X) = \frac 1n \sum_{i=1}^n \left( X_i - \bar X \right)^2 \\ = & \frac 1n \sum_{i=1}^n \left( X_i^2 - 2 X_i \bar X + \bar X^2 \right) \\ = & \frac 1n \sum_{i=1}^n X_i^2 - \frac {2 \bar X} n \sum_{i=1}^n X_i + \bar X^2 \\ = & \frac 1n \sum_{i=1}^n X_i^2 - \frac 1 {n^2} \left( \sum_{i=1}^n X_i \right)^2 \\ = & E \left( X^2 \right) - E^2 (X) \end{align*}
公式编号：
$$ 2a + 3b \leq 7 \tag 1 \label 1 $$
$$ 5a + 2b \leq 12 \tag 2 \label 2 $$
$\dfrac 3 {11} \times \eqref 1 + \dfrac 1 {11} \times \eqref 2$，得
$$ a + b \leq 3 \tag 3 \label 3 $$