【数学A】例題1.2.5：円順列・数珠順列（One More）★

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% 例題A1.2.5：円順列・数珠順列（One More）★
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1，2，3，4，5の数字が書かれた玉が1個ずつあるとき，次の問いに答えよ．

(1) これらの玉を円形に並べる方法は何通りあるか．

(2) これらの5個の玉から3個の玉を取り出して円形に並べる方法は何通りあるか．

(3) 1，2が隣り合うように円形に並べる方法は何通りあるか．

(4) これらの玉にひもを通し，首飾りを作る方法は何通りあるか．

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(1) 異なる5個の玉の円順列であるから，

\[
(5-1)!=4!=4\cdot 3\cdot 2\cdot 1=24\text{(通り)}
\]

(2) 異なる5個から3個選んだ円順列であるから，

\[
\frac{{}_5\mathrm{P}_3}{3}=\frac{5\cdot 4\cdot 3}{3}=20\text{(通り)}
\]

(3) 1，2をまとめて1つの玉と考えると，残りの3個と合わせた4個の円順列より，$(4-1)!$通り

そのそれぞれについて，1，2の並び方は，$2!$通り

よって，$(4-1)!\times 2!=12\text{(通り)}$

(4) 異なる5個の玉の円順列において，$(5-1)!$通りあるが，首飾りは裏返すことができる．

裏返すと同じものが2つずつできるから，

\[
\frac{(5-1)!}{2}=\frac{4\cdot 3\cdot 2\cdot 1}{2}=12\text{(通り)}
\]

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% 問題A1.2.5：円順列・数珠順列（One More）★
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A，B，C，D，E，Fの文字が書かれた玉が1個ずつあるとき，次の問いに答えよ．

(1) これらの玉を円形に並べる方法は何通りあるか．

(2) これらの6個の玉から4個の玉を取り出して円形に並べる方法は何通りあるか．

(3) D，Eが隣り合うように円形に並べる方法は何通りあるか．

(4) これらの玉にひもを通し，首飾りを作る方法は何通りあるか．

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(1) 異なる6個の玉の円順列であるから，

\[
(6-1)!=5!=5\cdot 4\cdot 3\cdot 2\cdot 1=120\text{(通り)}
\]

(2) 異なる6個から4個選んだ円順列であるから，

\[
\frac{{}_6\mathrm{P}_4}{4}=\frac{6\cdot 5\cdot 4\cdot 3}{4}=90(通り)
\]

(3) D，Eをまとめて1つの玉と考えると，残りの4個と合わせた異なる5個の玉の円順列より，$(5-1)!$通り

そのそれぞれについて，D，Eの並び方は，$2!$通り

よって，$(5-1)!\times 2!=48\text{(通り)}$

(4) 異なる6個の玉の円順列において，$(6-1)!$通りあるが，首飾りは裏返すことができる．

裏返すと同じものが2つずつできるから，

\[
\frac{(6-1)!}{2}=\frac{120}{2}=60\text{(通り)}
\]

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% 側注1
3つずつ重複するので，3で割る．

% 側注2
1，2と2，1の2通りある．

% 側注3
異なる$n$個の数珠順列は，$\frac{(n-1)!}{2}(\text{通り})$

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% 側注1
4つずつ重複するので，4で割る．

% 側注2
D，EとE，Dの2通りがある．

% 側注3
異なる$n$個の数珠順列は，$\frac{(n-1)!}{2}(\text{通り})$

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(2) 異なる3個の円順列と同様に，5個から3個選んで円形に並べる場合も，右の図のように重複する（同じものとなる）場合がある．

(3) 1，2をまとめて1つの玉として考えて，4個の円順列を求める．

(4) ひもを通し首飾りを作るとき，右の図のように円順列では異なる2通りとして考えたものが，裏返すと同じものになる．よって，円順列の場合の数を2で割ることで求めることができる．このような順列を数珠順列という．

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\[
\text{円順列の総数}=(n-1)!,\text{数珠順列の総数}=\frac{(n-1)!}{2}(n\geqq 3)
\]