問題の解答
検索用コード（LaTeX）
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% 例題A1.2.16：一部の文字の順序が定められた順列（One More）★★
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$\mathrm{holiday}$のすべての文字を1列に並べるとき，次の問いに答えよ．
(1) $\mathrm{o},\mathrm{i},\mathrm{a}$がこの順で現れる並び方は何通りあるか．
(2) $\mathrm{h}$が$\mathrm{l}$より左に，$\mathrm{d}$が$\mathrm{y}$より右に現れる並び方は何通りあるか．
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% 例題A1.2.16の解答（One More）★★
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(1) $\mathrm{o},\mathrm{i},\mathrm{a}$をすべて$\mathrm{X}$とおき，$\mathrm{X},\mathrm{X},\mathrm{X},\mathrm{h},\mathrm{l},\mathrm{d},\mathrm{y}$の7文字を1列に並べる順列の総数を求めればよい．
よって，求める総数は，
\[
\frac{7!}{3!}=\frac{7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{3\cdot 2\cdot 1}=840(\text{通り})
\]
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% 例題A1.2.16の別解
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7文字が入る7つの場所を考えて，o，i，aが入る場所をXとし，h，l，d，yが入る場所をYとする．
X，X，X，Y，Y，Y，Yの並び方の総数は，$\frac{7!}{3!4!}=35(\text{通り})$
Xにはo，i，aが順番に入るから，1通りであり，Yにはh，l，d，yが入るから，その順列は$4!=24(\text{通り})$
よって，求める総数は，$35\times 1\times 24=840(\text{通り})$
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% 例題A1.2.16の解答（One More）★★
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(2) $\mathrm{h},\mathrm{l}$を$\mathrm{X}$とおき，$\mathrm{d},\mathrm{y}$を$\mathrm{Y}$とおく．$\mathrm{X},\mathrm{X},\mathrm{Y},\mathrm{Y},\mathrm{o},\mathrm{i},\mathrm{a}$の7文字を1列に並べる順列の総数を求めればよい．
よって，求める総数は，
\[
\frac{7!}{2!2!}=\frac{7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{2\cdot 1\cdot 2\cdot 1}=1260(\text{通り})
\]
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% 問題A1.2.16：一部の文字の順序が定められた順列（One More）★★
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$\mathrm{sunlight}$のすべての文字を1列に並べるとき，次の問いに答えよ．
(1) $\mathrm{s},\mathrm{u},\mathrm{n}$がこの順で現れる並び方は何通りあるか．
(2) $\mathrm{s}$が$\mathrm{t}$より左に，$\mathrm{g}$が$\mathrm{h}$より右に現れる並び方は何通りあるか．
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% 問題A1.2.16の解答（One More）★★
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(1) $\mathrm{s},\mathrm{u},\mathrm{n}$をすべて$\mathrm{X}$とおき，$\mathrm{X},\mathrm{X},\mathrm{X},\mathrm{l},\mathrm{g},\mathrm{h},\mathrm{i},\mathrm{t}$の8文字を1列に並べる順列の総数を求めればよい．
よって，求める総数は，
\[
\frac{8!}{3!}=\frac{8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{3\cdot 2\cdot 1}=6720(\text{通り})
\]
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% 問題A1.2.16の別解
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8文字が入る8つの場所を考えて，$\mathrm{s},\mathrm{u},\mathrm{n}$が入る場所を$\mathrm{X}$とし，$\mathrm{l},\mathrm{i},\mathrm{g},\mathrm{h},\mathrm{t}$が入る場所を$\mathrm{Y}$とする．
$\mathrm{X},\mathrm{X},\mathrm{X},\mathrm{Y},\mathrm{Y},\mathrm{Y},\mathrm{Y},\mathrm{Y}$の並び方の総数は，$\frac{8!}{3!5!}=56(\text{通り})$
$\mathrm{X}$には$\mathrm{s},\mathrm{u},\mathrm{n}$が順番に入るから，1通りであり，$\mathrm{Y}$には$\mathrm{l},\mathrm{i},\mathrm{g},\mathrm{h},\mathrm{t}$が入るから，その順列は$5!=120(\text{通り})$
よって，求める総数は，$56\times 1\times 120=6720(\text{通り})$
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% 問題A1.2.16の解答（One More）★★
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(2) $\mathrm{s},\mathrm{t}$を$\mathrm{X}$とおき，$\mathrm{g},\mathrm{h}$を$\mathrm{Y}$とおく．$\mathrm{X},\mathrm{X},\mathrm{Y},\mathrm{Y},\mathrm{u},\mathrm{n},\mathrm{l},\mathrm{i}$の8文字を1列に並べる順列の総数を求めればよい．
よって，求める総数は，
\[
\frac{8!}{2!2!}=\frac{8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{2\cdot 1\cdot 2\cdot 1}=10080(\text{通り})
\]
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