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(A) Let $n_i$ be the number of balls chosen from box $i$,where $n_i \ge 2$ and $\sum_{i=1}^4 n_i = 10$. Since each box must contain at least one red a

Vedclass · Accepted Answer

$21816$

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(A) Let $n_i$ be the number of balls chosen from box $i$,where $n_i \ge 2$ and $\sum_{i=1}^4 n_i = 10$. Since each box must contain at least one red and one blue ball,the possible distributions of $(n_1, n_2, n_3, n_4)$ are permutations of $(4, 2, 2, 2)$ and $(3, 3, 2, 2)$. Case $I$: Distribution $(4, 2, 2, 2)$. Number of ways to choose $4$ balls from one box such that at least one red and one blue are chosen: $\binom{5}{4} - \binom{3}{4} - \binom{2}{4} = 5 - 0 - 0 = 5$. Number of ways to choose $2$ balls from one box such that at least one red and one blue are chosen: $\binom{3}{1} 	imes \binom{2}{1} = 3 	imes 2 = 6$. Total ways for this case: $\binom{4}{1} 	imes 5 	imes 6^3 = 4 	imes 5 	imes 216 = 4320$. Case $II$: Distribution $(3, 3, 2, 2)$. Number of ways to choose $3$ balls from one box such that at least one red and one blue are chosen: $\binom{5}{3} - \binom{3}{3} - \binom{2}{3} = 10 - 1 - 0 = 9$. Total ways for this case: $\binom{4}{2} 	imes 9^2 	imes 6^2 = 6 	imes 81 	imes 36 = 17496$. Total ways = $4320 + 17496 = 21816$.

Consider $4$ boxes,where each box contains $3$ red balls and $2$ blue balls. Assume that all $20$ balls are distinct. In how many different ways can $10$ balls be chosen from these $4$ boxes so that from each box at least one red ball and one blue ball are chosen?

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