【数学A】例題1.2.22：整数解の個数（One More）★★★★

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% 例題A1.2.22：整数解の個数（One More）★★★★
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次の式を満たす整数の組$(x,y,z)$は何通りあるか．

(1) $x+y+z=9,x\geqq 0,y\geqq 0,z\geqq 0$

(2) $x+y+z=9,x\geqq 1,y\geqq 1,z\geqq 1$

(3) $x+y+z\leqq 9,x\geqq 0,y\geqq 0,z\geqq 0$

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% 例題A1.2.22の解答（One More）★★★★
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(1) 求める組の総数は，9個の$\bigcirc$と2本の$|$の並び方を考えると，

\[
{}_{11}\mathrm{C}_{9}={}_{11}\mathrm{C}_2=55(\text{通り})
\]

(2) 9個の$\bigcirc$のうち，先に3個の$\bigcirc$を1個ずつ$x,y,z$に割り振ると考え，残りの6個の$\bigcirc$を$x,y,z$で割り振ればよい．つまり，求める組の総数は，6個の$\bigcirc$と2本の$|$の並び方を考えると，

\[
{}_8\mathrm{C}_6={}_8\mathrm{C}_2=28(\text{通り})
\]

(3) 求める組の総数は，9個の$\bigcirc$と3本の$|$を1列に並べて，

1本目の$|$より左側にある$\bigcirc$の個数は$x$の値，

1本目と2本目の$|$の間にある$\bigcirc$の個数は$y$の値，

2本目と3本目の$|$の間にある$\bigcirc$の個数は$z$の値

を表すとすると，求める組の総数は，

\[
{}_{12}\mathrm{C}_{9}={}_{12}\mathrm{C}_3=220(\text{通り})
\]

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% 問題A1.2.22：整数解の個数（One More）★★★★
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次の式を満たす整数の組$(x,y,z)$は何通りあるか．

(1) $x+y+z=8,x\geqq 0,y\geqq 0,z\geqq 0$

(2) $x+y+z=8,x\geqq 1,y\geqq 1,z\geqq 1$

(3) $x+y+z\leqq 8,x\geqq 0,y\geqq 0,z\geqq 0$

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% 問題A1.2.22の解答（One More）★★★★
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(1) 求める組の総数は，8個の$\bigcirc$と2本の$|$の並び方を考えると，

\[
{}_{10}\mathrm{C}_{8}={}_{10}\mathrm{C}_2=45(\text{通り})
\]

(2) 8個の$\bigcirc$のうち，先に3個の$\bigcirc$を1個ずつ$x,y,z$に割り振ると考え，残りの5個の$\bigcirc$を$x,y,z$で割り振ればよい．つまり，求める組の総数は，5個の$\bigcirc$と2本の$|$の並び方を考えると，${}_7\mathrm{C}_5={}_7\mathrm{C}_2=21(\text{通り})$

(3) 求める組の総数は，8個の$\bigcirc$と3本の$|$を1列に並べて，

1本目の$|$より左側にある$\bigcirc$の個数は$x$の値，

1本目と2本目の$|$の間にある$\bigcirc$の個数は$y$の値，

2本目と3本目の$|$の間にある$\bigcirc$の個数は$z$の値

を表すとすると，求める組の総数は，${}_{11}\mathrm{C}_{8}={}_{11}\mathrm{C}_3=165(\text{通り})$

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% 例題A1.2.22の側注：整数解の個数（One More）★★★★
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% 側注1
11個の場所から，$\bigcirc$が入る9個の場所を選ぶ．

% 側注2
$\frac{11!}{9!2!}$

% 側注3
$\bigcirc|\bigcirc\bigcirc\bigcirc|\bigcirc\bigcirc$のとき，$x=1+1,y=3+1,z=2+1$

% 側注4
$9-(x+y+z)=u$とおくと，$x+y+z+u=9,$ $x\geqq 0,y\geqq 0,z\geqq 0,u\geqq 0$となり，これらを満たす整数の組の個数と考えてもよい．

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% 問題A1.2.22の側注：整数解の個数（One More）★★★★
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% 側注1
10個の場所から，$\bigcirc$が入る8個の場所を選ぶ．

% 側注2
$\frac{10!}{8!2!}$

% 側注3
$\bigcirc|\bigcirc|\bigcirc\bigcirc\bigcirc$のとき，$x=1+1,y=1+1,z=3+1$

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% 例題A1.2.22の考え方・補足など：整数解の個数（One More）★★★★
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(1) 与えられた条件より，$x,y,z$は0以上の整数であるから，9個の$\bigcirc$を$x,y,z$に分けると考えると，$x,y,z$を合わせて9個選ぶ重複組合せと同じとなる．9個の$\bigcirc$と2本の$|$の合計11個の並び方の総数を求めればよい．

(1) 対応例

\[
\underbrace{\bigcirc\bigcirc\bigcirc\bigcirc}_{x(\text{4個})}|\underbrace{\bigcirc\bigcirc\bigcirc}_{y(\text{3個})}|\underbrace{\bigcirc\bigcirc}_{z(\text{2個})}\cdots(x,y,z)=(4,3,2)
\]

(2) $x,y,z$は1以上の整数（自然数）であるから，先に$x,y,z$に1個ずつ$\bigcirc$を割り振り，次に残りの6個の$\bigcirc$を$x,y,z$に割り振ることを考える．

(2)

\[
\underbrace{\bigcirc\bigcirc\bigcirc}_{x,y,z\text{に1個}}\underbrace{\bigcirc\bigcirc\bigcirc\bigcirc\bigcirc\bigcirc}_{\text{残りを}x,y,z\text{に割り振る}}
\]

(3) $x,y,z$についての式が不等式の場合には，右の図のように$x,y,z$，残りに対応させると考え，9個の$\bigcirc$と3本の$|$の合計12個の並び方の総数を求めればよい．

(3) 対応例

\[
\underbrace{\bigcirc\bigcirc}_{x}|\underbrace{\bigcirc\bigcirc\bigcirc}_{y}|\underbrace{\bigcirc\bigcirc}_{z}|\underbrace{\bigcirc\bigcirc}_{\text{残り}}\cdots(x,y,z)=(2,3,2)
\]

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和が一定となる0以上の整数の組の個数は，重複組合せで考える．