There are 9 balls to be placed in 9 boxes. If | vedclass

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(C) Let the $3$ small boxes be $S_1, S_2, S_3$ and the other $6$ boxes be $B_1, B_2, B_3, B_4, B_5, B_6$. There are $5$ balls that cannot fit into the

Vedclass · Accepted Answer

$17280$

Vedclass · Answer

(C) Let the $3$ small boxes be $S_1, S_2, S_3$ and the other $6$ boxes be $B_1, B_2, B_3, B_4, B_5, B_6$. There are $5$ balls that cannot fit into the $3$ small boxes. These $5$ balls must be placed in the $6$ larger boxes. The number of ways to arrange these $5$ balls in $6$ boxes is given by $^6P_5 = \frac{6!}{(6-5)!} = 6 	imes 5 	imes 4 	imes 3 	imes 2 = 720$. After placing these $5$ balls,there are $4$ balls remaining and $4$ boxes remaining (the $3$ small boxes and $1$ unused large box). The number of ways to arrange these $4$ balls in the remaining $4$ boxes is $4! = 24$. Total number of arrangements = $^6P_5 	imes 4! = 720 	imes 24 = 17280$.

There are $9$ balls to be placed in $9$ boxes. If $5$ of the balls cannot fit into $3$ specific small boxes,find the number of ways of arranging one ball in each of the boxes.

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