# 統計学のための数学入門30講をひたすら解いてみる(第3講)

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# 第3講 順列・組合せと2項定理・多項定理

## 3.1 順列と組合せ

### (3.1)

\begin{align}
{}_n P_k &= n(n-1)(n-2) \cdots (n-k+2)(n-k+1) \\
&= n(n-1)(n-2) \cdots (n-(k-2))(n-(k-1)) \\
&= \frac{n(n-1)(n-2) \cdots (n-(k-2))(n-(k-1))(n-k) \cdots 2 \cdot 1}{(n-k)(n-(k+1)) \cdots 2 \cdot 1} \\
&= \frac{n!}{(n-k)!}
\end{align}


### (3.3)

\begin{align}
{}_n P_n &= \frac{n!}{(n-n)!} \\
&= \frac{n!}{0!} = n!
\end{align}

\begin{align}
{}_n P_0 &= \frac{n!}{(n-0)!} \\
&= \frac{n!}{n!} = 1
\end{align}


### (3.5)

\begin{align}
{}_n C_n &= \frac{n!}{(n-n)!n!} \\
&= \frac{n!}{0!n!} = 1
\end{align}

\begin{align}
{}_n C_0 &= \frac{n!}{(n-0)!0!} \\
&= \frac{n!}{n!0!} = 1
\end{align}


### 問題3.1(1)

\begin{align}
{}_8 P_3 &= \frac{8!}{5!} \\
&= 8 \cdot 7 \cdot 6 \\
&= 336
\end{align}


### 問題3.1(2)

\begin{align}
{}_8 C_3 &= \frac{8!}{5!3!} \\
&= \frac{336}{3!} \\
&= \frac{336}{3 \cdot 2 \cdot 1} \\
&= 56
\end{align}


### 問題3.1(3)

\begin{align}
{}_8 C_5 &= {}_8 C_{(8-3)!} \\
&= {}_8 C_3 \\
&= 56
\end{align}


### 問題3.2(1)

\begin{align}
(1+1)^6 &= {}_6 C_0 \cdot 1^6 \cdot 1^0 + {}_6 C_1 \cdot 1^5 \cdot 1^1 + \cdots + {}_6 C_5 \cdot 1^1 \cdot 1^5 + {}_6 C_6 \cdot 1^0 \cdot 1^6 \\
&= 2^6 = 64
\end{align}


### 問題3.2(2)

\begin{align}
\{1+(-1)\}^6 &= {}_6 C_0 \cdot 1^6 \cdot (-1)^0 + {}_6 C_1 \cdot 1^5 \cdot (-1)^1 + \cdots + {}_6 C_5 \cdot 1^1 \cdot (-1)^5 + {}_6 C_6 \cdot 1^0 \cdot (-1)^6 \\
&= {}_6 C_0 - {}_6 C_1 + \cdots - {}_6 C_5 + {}_6 C_6 \\
&= 0^6 = 0
\end{align}