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

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(C) Let the $9$ boxes be $B_1, B_2, \dots, B_9$ and the $3$ small boxes be $B_1, B_2, B_3$. There are $5$ balls (say $b_1, b_2, b_3, b_4, b_5$) that c

Vedclass · Accepted Answer

$17280$

Vedclass · Answer

(C) Let the $9$ boxes be $B_1, B_2, \dots, B_9$ and the $3$ small boxes be $B_1, B_2, B_3$. There are $5$ balls (say $b_1, b_2, b_3, b_4, b_5$) that cannot fit into these $3$ small boxes. These $5$ balls must be placed in the remaining $9 - 3 = 6$ boxes. The number of ways to arrange these $5$ balls in $6$ boxes is given by $^6P_5$. After placing these $5$ balls,we are left with $9 - 5 = 4$ balls and $9 - 5 = 4$ boxes. The number of ways to arrange the remaining $4$ balls in the remaining $4$ boxes is $4!$. Total number of arrangements = $^6P_5 	imes 4!$. $^6P_5 = \frac{6!}{(6-5)!} = 6 	imes 5 	imes 4 	imes 3 	imes 2 = 720$. $4! = 4 	imes 3 	imes 2 	imes 1 = 24$. Total arrangements = $720 	imes 24 = 17280$.

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 to arrange exactly one ball in each of the boxes.

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