2021학년도 09월 나형 29번

29. 흰 공 $4$$4$개와 검은 공 $6$$6$개를 세 상자 $A,B,C$$A,~B,~C$에 남김없이 나누어 넣을 때, 각 상자에 공이 $2$$2$개 이상씩 들어가도록 나누어 넣는 경우의 수를 구하시오. (단, 같은 색 공끼리는 서로 구별하지 않는다.)

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i. 정리

• $w:4,b:6$$w: 4,~b:6$
• $A,B,C$$A,~B,~C$에 두 개 이상씩

ii. 생각

• 아무래도 여사건이겠지? 후후

  • $2$$2$개가 $0$$0$
  • $1$$1$개가 $0$$0$

  ....이건 아닌 듯...;;;

• 나눠서 넣기 힘드니 반대로 생각해볼까?

  $(A,B,C)$$(A,~B,~C)$가 받는 순서쌍을 구하면,

  $(2,2,6),(2,3,5),(2,4,4),(3,3,4)$$(2,~2,~6),~(2,~3,~5),~(2,~4,~4),~(3,~3,~4)$

  흠..

  $w$$w$를 정하면, $b$$b$가 정해지네?

• 경우의 수를 구하자.

  • $(2,2,6)$$(2,~2,~6)$인 경우

    	$A$$A$	$B$$B$	$C$$C$
    $\begin{array}{c}w\\ b\end{array}$$w\\ b$	$2$$2$	$2$$2$	$6$$6$
    $\begin{array}{c}w\\ b\end{array}$$w\\ b$	$\begin{array}{c}0\\ 2\end{array}$$0\\2$	$\begin{array}{c}0\\ 2\end{array}$$0\\2$	$\begin{array}{c}4\\ 2\end{array}$$4\\2$
    $\begin{array}{c}w\\ b\end{array}$$w\\ b$	$\begin{array}{c}0\\ 2\end{array}$$0\\2$	$\begin{array}{c}1\\ 1\end{array}$$1\\1$	$\begin{array}{c}3\\ 3\end{array}$$3\\3$
    $\begin{array}{c}w\\ b\end{array}$$w\\ b$	$\begin{array}{c}0\\ 2\end{array}$$0\\2$	$\begin{array}{c}2\\ 0\end{array}$$2\\0$	$\begin{array}{c}2\\ 4\end{array}$$2\\4$
    $\begin{array}{c}w\\ b\end{array}$$w\\ b$	$\begin{array}{c}1\\ 1\end{array}$$1\\1$	$\begin{array}{c}0\\ 2\end{array}$$0\\2$	$\begin{array}{c}3\\ 3\end{array}$$3\\3$
    $\begin{array}{c}w\\ b\end{array}$$w\\ b$	$\begin{array}{c}1\\ 1\end{array}$$1\\1$	$\begin{array}{c}1\\ 1\end{array}$$1\\1$	$\begin{array}{c}2\\ 4\end{array}$$2\\4$
    $\begin{array}{c}w\\ b\end{array}$$w\\ b$	$\begin{array}{c}1\\ 1\end{array}$$1\\1$	$\begin{array}{c}2\\ 0\end{array}$$2\\0$	$\begin{array}{c}1\\ 3\end{array}$$1\\3$
    $\begin{array}{c}w\\ b\end{array}$$w\\ b$	$\begin{array}{c}2\\ 0\end{array}$$2\\0$	$\begin{array}{c}0\\ 2\end{array}$$0\\2$	$\begin{array}{c}2\\ 4\end{array}$$2\\4$
    $\begin{array}{c}w\\ b\end{array}$$w\\ b$	$\begin{array}{c}2\\ 0\end{array}$$2\\0$	$\begin{array}{c}1\\ 1\end{array}$$1\\1$	$\begin{array}{c}1\\ 5\end{array}$$1\\5$
    $\begin{array}{c}w\\ b\end{array}$$w\\ b$	$\begin{array}{c}2\\ 0\end{array}$$2\\0$	$\begin{array}{c}2\\ 0\end{array}$$2\\0$	$\begin{array}{c}0\\ 6\end{array}$$0\\6$

    뭐 문제 풀때는 $w$$w$만 쓰면 되겠지?

    이를 배열하는 경우는 $3$$3$가지

    $\therefore 9\times 3=27$$\therefore~ 9\times 3=27$

  • $(2,3,5)$$(2,~3,~5)$의 경우

    $w$$w$만 쓰자.

    $(0,0,4),(0,1,3),(0,2,2),(0,3,1),$$(0,~0,~4),~(0,~1,~3),~(0,~2,~2),~(0,~3,~1),$어우 많겠는데?

    여사건으로 풀자.

    $(3,0,1),(3,1,0),(0,4,0),(4,0,0)$$(3,~0,~1),~(3,~1,~0),~(0,~4,~0),~(4,~0,~0)$

    ${}_3{H}_4-4=15-4=11$${}_3H_4-4=15-4=11$

    이를 배열하는 경우는 $3!$$3!$

    $\therefore 11\times 3!=66$$\therefore~ 11\times 3!=66$

  • $(2,4,4)$$(2,~4,~4)$

    $(3,0,1),(3,1,0),(4,0,0)$$(3,~0,~1),~(3,~1,~0),~(4,~0,~0)$

    ${}_3{H}_4-3=12$${}_3H_4-3=12$

    배열하는 경우는 $3$$3$

    $\therefore 12\times 3=36$$\therefore~ 12\times 3=36$

  • $(3,3,4)$$(3,~3,~4)$

    $(4,0,0),(0,4,0)$$(4,~0,~0),~(0,~4,~0)$

    ${}_3{H}_4-2=13$${}_3H_4-2=13$

    배열하는 경우는 $3$$3$

    $\therefore 13\times 3$$\therefore~ 13\times 3$

$\therefore 27+66+36+39=168$$\therefore~ 27+66+36+39=168$