In a circus, there are 10 cages to accommodat | vedclass

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(B) Let the $10$ animals be $A_1, A_2, \dots, A_{10}$ and the $10$ cages be $C_1, C_2, \dots, C_{10}$.Let $C_1, C_2, C_3, C_4$ be the $4$ small cages.

Vedclass · Accepted Answer

$86400$

Vedclass · Answer

(B) Let the $10$ animals be $A_1, A_2, \dots, A_{10}$ and the $10$ cages be $C_1, C_2, \dots, C_{10}$.Let $C_1, C_2, C_3, C_4$ be the $4$ small cages.Let $A_1, A_2, A_3, A_4, A_5$ be the $5$ animals that cannot enter the small cages.These $5$ animals must be placed in the remaining $6$ cages $(C_5, C_6, C_7, C_8, C_9, C_{10})$.The number of ways to arrange these $5$ animals in $6$ cages is given by the permutation formula $^6P_5 = \frac{6!}{(6-5)!} = 6! = 720$.After placing these $5$ animals, we have $5$ animals remaining and $5$ cages remaining (the $4$ small cages plus the $1$ cage left over from the $6$ large cages).The number of ways to arrange the remaining $5$ animals in the remaining $5$ cages is $5! = 120$.Therefore, the total number of ways = $^6P_5 	imes 5! = 720 	imes 120 = 86400$.

In a circus, there are $10$ cages to accommodate $10$ animals. Out of these, $4$ cages are so small that $5$ specific animals cannot enter them. In how many ways is it possible to accommodate the $10$ animals in these $10$ cages?

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