【数学A】例題1.2.12：長方形の個数（One More）★★

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% 例題A1.2.12：長方形の個数（One More）★★
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縦の長さが5，横の長さが7の長方形を，右の図のように縦を5等分，横を7等分に区切るとする．このとき，この図形に含まれる線分を辺とする次の図形の個数を求めよ．

(1) 長方形の個数

(2) 正方形の個数

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% 例題A1.2.12の解答（One More）★★
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(1) 8本の縦線から2本を選び，6本の横線から2本を選ぶと1個の長方形が定まる．

よって，長方形の個数は，

\[
{}_8\mathrm{C}_2\times{}_6\mathrm{C}_2=28\times 15=420(\text{個})
\]

(2) この図形には，1辺が1，2，3，4，5の5種類の正方形が含まれている．

1辺が1の正方形は，縦線，横線から幅が1の2本を選ぶと，1個の正方形が定まる．

したがって，縦線7通り，横線5通りより，$7\times 5=35(\text{個})$

同様に，1辺が2の正方形は，縦線6通り，横線4通りより，$6\times 4=24(\text{個})$

1辺が3の正方形は，縦線5通り，横線3通りより，$5\times 3=15(\text{個})$

1辺が4の正方形は，縦線4通り，横線2通りより，$4\times 2=8(\text{個})$

1辺が5の正方形は，縦線3通り，横線1通りより，$3\times 1=3(\text{個})$

よって，正方形の個数は，

\[
35+24+15+8+3=85(\text{個})
\]

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縦の長さが4，横の長さが7の長方形を，右の図のように縦を4等分，横を7等分に区切るとする．このとき，この図形に含まれる線分を辺とする次の図形の個数を求めよ．

(1) 長方形の個数

(2) 正方形の個数

(3) 長方形であって正方形ではないもの

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(1) 8本の縦線から2本を選び，5本の横線から2本を選ぶと1個の長方形が定まる．

よって，長方形の個数は，${}_8\mathrm{C}_2\times{}_5\mathrm{C}_2=28\times 10=280(\text{個})$

(2) この図形には，1辺が1，2，3，4の4種類の正方形が含まれている．

1辺が1の正方形は，縦線，横線から幅が1の2本を選ぶと，1個の正方形が定まる．

したがって，縦線7通り，横線4通りより，$7\times 4=28(\text{個})$

同様に，1辺が2の正方形は，縦線6通り，横線3通りより，$6\times 3=18(\text{個})$

1辺が3の正方形は，縦線5通り，横線2通りより，$5\times 2=10(\text{個})$

1辺が4の正方形は，縦線4通り，横線1通りより，$4\times 1=4(\text{個})$

よって，正方形の個数は，$28+18+10+4=60(\text{個})$

(3) (1)，(2)より，長方形の個数は280個，正方形の個数は60個である．

よって，長方形であって正方形ではないものの個数は，$280-60=220(\text{個})$

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% 側注1
積の法則を用いる．

% 側注2
隣り合う2本を考える．

% 側注3
積の法則を用いる．

% 側注4
間隔が2である2本の縦線と2本の横線（1辺が2の正方形）の選び方はそれぞれ6通り，4通りとなる．間隔が3，4，5の場合も同様に考える．

% 側注5
和の法則を用いる．

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% 問題A1.2.12の側注：長方形の個数（One More）★★
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% 側注1
積の法則を用いる．

% 側注2
隣り合う2本を考える．

% 側注3
積の法則を用いる．

% 側注4
和の法則を用いる．

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% 例題A1.2.12の考え方・補足など：長方形の個数（One More）★★
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(1) 右の図のように，縦線2本と横線2本の線分を選べば，1個の長方形を定めることができる．なお，正方形も長方形に含まれることに注意すること．

(2) 長方形のときとは異なり，縦線，横線からそれぞれ同じ間隔の2本を選ばなければならない．組合せを用いることが難しいため，積の法則，和の法則を用いて数え上げる．なお，例えば間隔が2である2本の縦線と2本の横線（1辺の長さが2の正方形）は，右の図のようになる．

(1) 1個の長方形が定まる

(2) 間隔が2である2本の縦線と2本の横線の例