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

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(B) There are $10$ animals and $10$ cages. Let the set of animals be $A$ and the set of cages be $C$. $4$ cages are small,meaning $5$ specific animals

Vedclass · Accepted Answer

$86400$

Vedclass · Answer

(B) There are $10$ animals and $10$ cages. Let the set of animals be $A$ and the set of cages be $C$. $4$ cages are small,meaning $5$ specific animals cannot enter them. Let $S$ be the set of $5$ animals that cannot enter the $4$ small cages. These $5$ animals must be placed in the remaining $10 - 4 = 6$ large cages. The number of ways to arrange these $5$ animals in $6$ cages is given by $^6P_5 = \frac{6!}{(6-5)!} = 6 	imes 5 	imes 4 	imes 3 	imes 2 = 720$. After placing these $5$ animals,we have $5$ animals remaining and $5$ cages remaining (the $4$ small cages plus the $1$ large cage left over). The number of ways to arrange the remaining $5$ animals in the remaining $5$ cages is $5! = 120$. Total number of ways = $^6P_5 	imes 5! = 720 	imes 120 = 86400$.

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

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