統計学のための数学

場合の数

順列

\begin{aligned} _{n} P_{k} & = n (n - 1) (n - 2) \dots (n - (k - 1)) \\ = \frac{n!}{(n - k)!} \end{aligned}

$\begin{align} {}_{n}P_{k} &= n(n-1)(n-2)\dots(n-(k-1)) \\ &= \frac{n!}{(n-k)!} \end{align}$

_{n} P_{0} = 1

$\begin{equation} {}_{n}P_{0} = 1 \end{equation}$

組み合せ

\begin{aligned} _{n} C_{k} & = \frac{_{n} P_{k}}{k!} \\ = \frac{n!}{k! (n - k)!} \end{aligned}

$\begin{align} {}_{n}C_{k} &= \frac{_{n}P_{k}}{k!} \\ &= \frac{n!}{k!\,(n-k)!} \end{align}$

_{n} C_{0} = 1

$\begin{equation} {}_{n}C_{0} = 1 \end{equation}$

_{n} C_{k} =_{n} C_{n - k}

$\begin{equation} {}_{n}C_{k} = {}_{n}C_{n - k} \end{equation}$

重複組み合せ

区別できる $n$ 個のものから $k$ 個(必ずしも相異なるとは限らない)選んでできる組み合せの総数は，

\begin{aligned} _{n} H_{k} & =_{n + k - 1} C_{k} \\ = \frac{n (n + 1) \dots (n + k - 1)}{k!} \end{aligned}

$\begin{align} {}_{n}H_{k} &= {}_{n+k-1}C_{k} \\ &= \frac{n(n+1)\cdots(n+k-1)}{k!} \end{align}$

n + k - 1

$n+k-1$ 個の区別できる箱から

k

$k$ 個選んでボールを入れる組み合わせと言い換えることができる．

例題

x + y + z = 10, x, y, z \geq 0, x, y, z \in R

$\begin{equation} x+y+z=10,~~x,y,z \ge 0,~~x,y,z \in \mathbb{R} \end{equation}$

の解の個数を求めなさい．

[解答]

_{3} H_{10} =_{10 + 3 - 1} C_{10} =_{12} C_{2} = 66

$\begin{equation} {}_{3}H_{10} = {}_{10+3-1}C_{10} = {}_{12}C_{2} = 66 \end{equation}$

二重階乗

$k \in N$ のとき，

k!! = {\begin{cases} 2^{m} m! & k = 2 m (m = 1, 2, \dots) \\ \frac{2 m!}{2^{m} m!} & k = 2 m - 1 (m = 1, 2, \dots) \end{cases}

$\begin{equation} k!! = \begin{cases} 2^m\,m! & k = 2m~~(m = 1, 2, \ldots) \\ \dfrac{2m!}{2^m\,m!} & k = 2m-1~~(m = 1, 2, \ldots) \end{cases} \end{equation}$

証明

$k = 2m$ のとき，

\begin{aligned} k!! & = (2 m) \times (2 m - 2) \times (2 m - 4) \times \dots \times 2 \\ = 2 m \times 2 (m - 1) \times 2 (m - 2) \times \dots \times 2 \cdot 1 \\ = 2^{m} m! \end{aligned}

$\begin{align} k!! &= (2m) \times (2m - 2) \times (2m - 4) \times \cdots \times 2 \\ &= 2m \times 2(m - 1) \times 2(m - 2) \times \cdots \times 2 \cdot 1 \\ &= 2^m\,m! \end{align}$

k = 2 m - 1

$k = 2m - 1$ のとき，

\begin{aligned} 2 m! & = (2 m)!! \times (2 m - 1)!! \\ = (2 m)!! \times k!! \\ = 2^{m} m! \times k!! \end{aligned}

$\begin{align} 2m! &= (2m)!! \times (2m-1)!! \\ &= (2m)!! \times k!! \\ &= 2^m\,m! \times k!! \end{align}$

∴ k!! = \frac{2 m!}{2^{m} m!}

$\begin{equation} \therefore~k!! = \dfrac{2m!}{2^m\,m!} \end{equation}$

補足

$k = 2m+1$ のとき，

k!! = \frac{Γ (m + 1 / 2)}{Γ (1 / 2)} 2^{m}

$\begin{equation} k!! = \dfrac{\varGamma\left(m + 1/2\right)}{\varGamma\left(1/2\right)}2^m \end{equation}$

ともかける．