In a circus,there are 10 cages for 10 animals | vedclass

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(B) Let the $5$ animals that cannot enter the $4$ small cages be called 'large animals' and the other $5$ animals be 'small animals'.There are $6$ lar

Vedclass · Accepted Answer

$86400$

Vedclass · Answer

(B) Let the $5$ animals that cannot enter the $4$ small cages be called 'large animals' and the other $5$ animals be 'small animals'.There are $6$ large cages and $4$ small cages.The $5$ large animals must be placed in the $6$ large cages. This can be done in $_6P_5$ ways.After placing the $5$ large animals,we have $1$ large cage left and $4$ small cages left,totaling $5$ cages for the remaining $5$ small animals.The $5$ small animals can be placed in these $5$ cages in $5!$ ways.Therefore,the total number of ways is $_6P_5 	imes 5!$.$_6P_5 = \frac{6!}{1!} = 6 	imes 5 	imes 4 	imes 3 	imes 2 = 720$.$5! = 120$.Total ways = $720 	imes 120 = 86400$.

In a circus,there are $10$ cages for $10$ animals. $4$ cages are so small that $5$ out of the $10$ animals cannot enter them. In how many ways can the $10$ animals be placed in the $10$ cages?

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