【数学A】例題1.2.3：条件付きの順列2（One More）★★

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% 例題A1.2.3：条件付きの順列2（One More）★★
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大人5人，子供3人の合計8人が1列に並ぶ．このとき，次の条件を満たす並び方は何通りあるか．

(1) 並び方の総数

(2) 両端が大人である

(3) 少なくとも一方の端が子供である

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(1) 8人が1列に並ぶ順列であるから，並び方の総数は，
\[
{}_8\mathrm{P}_8=8!=8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1=40320\text{(通り)}
\]

(2) 両端が大人である並び方は，${}_5\mathrm{P}_2=5\cdot 4=20(\text{通り})$

残りの6人を並べる順列は，$6!=6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1=720(\text{通り})$

よって，両端が大人である並び方は，

\[
20\times 720=14400\text{(通り)}
\]

(3) 少なくとも一方の端が子供である並び方は，全体から両端が大人である並び方を引いたものである．

よって，(1)，(2)より，少なくとも一方の端が子供である並び方は，

\[
40320-14400=25920\text{(通り)}
\]

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% 問題A1.2.3：条件付きの順列2（One More）★★
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男性4人，女性4人の合計8人が1列に並ぶ．このとき，次の条件を満たす並び方は何通りあるか．

(1) 並び方の総数

(2) 両端が男性である

(3) 少なくとも一方の端が女性である

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% 問題A1.2.3の解答（One More）★★
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(1) 8人が1列に並ぶ順列であるから，並び方の総数は，
\[
{}_8\mathrm{P}_8=8!=8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1=40320\text{(通り)}
\]

(2) 両端が男性である並び方は，${}_4\mathrm{P}_2=4\cdot 3=12(\text{通り})$

残りの6人を並べる順列は，$6!=6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1=720(\text{通り})$

よって，両端が男性である並び方は，

\[
12\times 720=8640\text{(通り)}
\]

(3) 少なくとも一方の端が女性である並び方は，全体から両端が男性である並び方を引いたものである．

よって，(1)，(2)より，少なくとも一方の端が女性である並び方は，

\[
40320-8640=31680\text{(通り)}
\]

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% 例題A1.2.3の側注：条件付きの順列2（One More）★★
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% 側注1
大人5人から2人が1列に並ぶ順列を考える（両端には右端と左端があるから，単に選ぶだけではなく，順序も考える）．

% 側注2
$n(A)=n(U)-n(\overline{A})$

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% 問題A1.2.3の側注：条件付きの順列2（One More）★★
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% 側注1
男性4人から2人が1列に並ぶ順列を考える（両端には右端と左端があるから，単に選ぶだけではなく，順序も考える）．

% 側注2
$n(A)=n(U)-n(\overline{A})$

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% 例題A1.2.3の考え方・補足など：条件付きの順列2（One More）★★
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(2) 先に両端に大人を並べ，次に間に入る6人の並び方を考える．

(3) 少なくとも一方の端が子供である並び方は，両端が子供，左端が子供で右端が大人，左端が大人で右端が子供の3つの場合がある．場合分けを用いて求めようとすると手間が掛かるので，並び方の総数から，両端が大人である並び方を引くことを考えるとよい．

$($少なくとも一方の端が子供である$)$ $=($並び方の総数$)-$ $($両端が大人である$)$