現在,One Moreの数学II・B版を作成中です!

【数学A】例題1.2.15:同じものを含む順列(One More)★

【数学A】例題1.2.15:同じものを含む順列(One More)
【数学A】例題1.2.15:同じものを含む順列の例題ページ
問題の解答

【数学A】問題1.2.15:同じものを含む順列の解答
検索用コード(LaTeX)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 例題A1.2.15:同じものを含む順列(One More)★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

次の問いに答えよ.

(1) $a,a,b,b,b,b$の6文字を1列に並べる順列は何通りあるか.

(2) 赤色のカード3枚,青色のカード3枚,緑色のカード1枚の合計7枚を1列に並べる順列は何通りあるか.

(3) 赤玉5個と白玉4個の合計9個を1列に並べる順列は何通りあるか.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 例題A1.2.15の解答(One More)★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

(1) 2個の$a$と4個の$b$を含む6個の順列であるから,

\[
\frac{6!}{2!4!}=\frac{6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{2\cdot 1\cdot 4\cdot 3\cdot 2\cdot 1}=15\text{(通り)}
\]

(2) 3枚の赤色のカードと3枚の青色のカード,1枚の緑色のカードを含む7枚の順列であるから,

\[
\frac{7!}{3!3!1!}=\frac{7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{3\cdot 2\cdot 1\cdot 3\cdot 2\cdot 1\cdot 1}=140\text{(通り)}
\]

(3) 5個の赤玉と4個の白玉を含む9個の順列であるから,

\[
\frac{9!}{5!4!}=\frac{9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{5\cdot 4\cdot 3\cdot 2\cdot 1\cdot 4\cdot 3\cdot 2\cdot 1}=126\text{(通り)}
\]

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 例題A1.2.15の別解
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

(1) ${}_6\mathrm{C}_2=15$\text{(通り)}

(2) ${}_7\mathrm{C}_3\times{}_4\mathrm{C}_3\times{}_1\mathrm{C}_1=35\times 4\times 1=140$\text{(通り)}

(3) ${}_9\mathrm{C}_5=126$\text{(通り)}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 問題A1.2.15:同じものを含む順列(One More)★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

次の問いに答えよ.

(1) $x,x,x,y,y,z$の6文字を1列に並べる順列は何通りあるか.

(2) 青玉6個と緑玉3個の合計9個を1列に並べる順列は何通りあるか.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 問題A1.2.15の解答(One More)★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

(1) 3個の$x$,2個の$y$,1個の$z$を含む6個の順列であるから,

\[
\frac{6!}{3!2!1!}=\frac{6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{3\cdot 2\cdot 1\cdot 2\cdot 1\cdot 1}=60\text{(通り)}
\]

(2) 6個の青玉と3個の緑玉を含む9個の順列であるから,

\[
\frac{9!}{6!3!}=\frac{9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1\cdot 3\cdot 2\cdot 1}=84\text{(通り)}
\]

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 問題A1.2.15の別解
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

(1) ${}_6\mathrm{C}_3\times{}_3\mathrm{C}_2=60$\text{(通り)}

(2) ${}_9\mathrm{C}_6=84$\text{(通り)}

あわせて読みたい
【数学A】1章:場合の数(基本事項) 検索用コード(LaTeX) 本文・解答側注 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % 基本事項A1.1.1:集合(One More) %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%...
あわせて読みたい
【数学A】1章:場合の数(節末問題・章末問題) 節末A1.1.1〜A1.1.5の解答 節末A1.1.1節末A1.1.2節末A1.1.3節末A1.1.4節末A1.1.5 リンク(関連例題) https://onemath.net/onemorea-reidai1-1-2 https://onemath.net/o...
目次