数学A

階乗

$n! = n(n-1)(n-2) \cdots 3 \cdot 2 \cdot 1$

$$
n! = n(n-1)(n-2) \cdots 3 \cdot 2 \cdot 1
$$

${}_n\mathrm{P}_r = n(n-1)(n-2) \cdots (n-r+1) = \dfrac{n!}{(n-r)!}$

$$
{}_n\mathrm{P}_r = n(n-1)(n-2) \cdots (n-r+1) = \dfrac{n!}{(n-r)!}
$$

$(n-1)! \ \text{通り}$

$$
(n-1)! \ \text{通り}
$$

$\dfrac{(n-1)!}{2} \ \text{通り}$

$$
\dfrac{(n-1)!}{2} \ \text{通り}
$$

${}_n\mathrm{C}_r = \dfrac{{}_n\mathrm{P}_r}{r!} = \dfrac{n!}{r!(n-r)!}$

$$
{}_n\mathrm{C}_r = \dfrac{{}_n\mathrm{P}_r}{r!} = \dfrac{n!}{r!(n-r)!}
$$

${}_n\mathrm{C}_r = {}_n\mathrm{C}_{n-r}$

$$
{}_n\mathrm{C}_r = {}_n\mathrm{C}_{n-r}
$$

${}_n\mathrm{C}_r = {}_{n-1}\mathrm{C}_{r-1} + {}_{n-1}\mathrm{C}_r$

$$
{}_n\mathrm{C}_r = {}_{n-1}\mathrm{C}_{r-1} + {}_{n-1}\mathrm{C}_r
$$

$k \cdot {}_n\mathrm{C}_k = n \cdot {}_{n-1}\mathrm{C}_{k-1}$

$$
k \cdot {}_n\mathrm{C}_k = n \cdot {}_{n-1}\mathrm{C}_{k-1}
$$

$P(A) = \dfrac{n(A)}{n(U)}$

$$
P(A) = \dfrac{n(A)}{n(U)}
$$

$P(\overline{A}) = 1 - P(A)$

$$
P(\overline{A}) = 1 - P(A)
$$

$P_A(B) = \dfrac{P(A \cap B)}{P(A)}$

$$
P_A(B) = \dfrac{P(A \cap B)}{P(A)}
$$

$a \equiv b \quad (\bmod \ m)$

$$
a \equiv b \quad (\bmod \ m)
$$

$a^{p-1} \equiv 1 \quad (\bmod \ p)$

$$
a^{p-1} \equiv 1 \quad (\bmod \ p)
$$