【数学A】例題1.1.9：出る目の総数を用いる場合の数（One More）★★

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% 例題A1.1.9：出る目の総数を用いる場合の数（One More）★★
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大，中，小3個のさいころを投げるとき，目の積が3の倍数となる場合は何通りあるか．

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% 例題A1.1.9の解答（One More）★★
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さいころの出る目の総数は，

\[
6\times 6\times 6=216(\text{通り})
\]

さいころの目の積が3の倍数となるのは，3個のさいころの目のうち少なくとも1つが3または6である場合である．

3個のさいころの目がすべて3または6以外である場合の数は，

\[
4\times 4\times 4=64(\text{通り})
\]

よって，求める場合の数は，$216-64=152(\text{通り})$

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大，中，小3個のさいころを投げるとき，目の積が5の倍数となる場合は何通りあるか．

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% 問題A1.1.9の解答（One More）★★
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さいころの出る目の総数は，$6\times 6\times 6=216(\text{通り})$

さいころの目の積が5の倍数となるのは，3個のさいころの目のうち少なくとも1つが5である場合である．

3個のさいころの目がすべて5以外である場合の数は，

\[
5\times 5\times 5=125(\text{通り})
\]

よって，求める場合の数は，$216-125=91(\text{通り})$

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% 例題A1.1.9の側注：出る目の総数を用いる場合の数（One More）★★
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% 側注1
3個とも1，2，4，5の目が出る場合の数を考える．

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% 問題A1.1.9の側注：出る目の総数を用いる場合の数（One More）★★
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% 側注1
3個とも1，2，3，4，6の目が出る場合の数を考える．

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% 例題A1.1.9の考え方・補足など：出る目の総数を用いる場合の数（One More）★★
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% 考え方
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目の積が3の倍数である場合の数を，場合分けを用いて求めようとすると手間が掛かる．そこで，出る目の総数から目の積が3の倍数ではないものを引いた，$($目の積が3の倍数$)=($出る目の総数$)-($目の積が3の倍数ではない$)$を考えるとよい．ここで，目の積が3の倍数にならないのは，3個とも1，2，4，5の目（4通り）の場合である．

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目の積が3の倍数になる場合の数を場合分けを用いて求めると，次のようになる．

(i) 3つの目がすべて3の倍数のとき，$2^3=8(\text{通り})$

(ii) 2つの目が3の倍数で，残り1つの目が3の倍数ではないとき，$(2^2\times 4)\times 3=48(\text{通り})$

(iii) 1つの目が3の倍数で，残り2つの目が3の倍数ではないとき，$(2\times 4^2)\times 3=96(\text{通り})$

よって，(i)〜(iii)より，求める場合の数は，$8+48+96=152(\text{通り})$

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全体から引いたものを考えると，楽に求めることができる．