Five balls of different colors are to be plac | vedclass

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(A) Let the number of balls in the three boxes be $(n_1, n_2, n_3)$ such that $n_1 + n_2 + n_3 = 5$ and $n_i \ge 1$.The possible partitions of $5$ int

Vedclass · Accepted Answer

$150$

Vedclass · Answer

(A) Let the number of balls in the three boxes be $(n_1, n_2, n_3)$ such that $n_1 + n_2 + n_3 = 5$ and $n_i \ge 1$.The possible partitions of $5$ into $3$ parts are $(3, 1, 1)$ and $(2, 2, 1)$.Case $1$: Partition $(3, 1, 1)$.The number of ways to distribute the balls is $\frac{5!}{3!1!1!} 	imes \frac{3!}{2!} = 20 	imes 3 = 60$.Case $2$: Partition $(2, 2, 1)$.The number of ways to distribute the balls is $\frac{5!}{2!2!1!} 	imes \frac{3!}{2!} = 30 	imes 3 = 90$.Total number of ways = $60 + 90 = 150$.

Five balls of different colors are to be placed in three boxes of different sizes,where each box can hold all five balls. In how many ways can the balls be placed such that no box remains empty?

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