In a circus there are ten cages for accommodating ten animals. Out of these four cages are so small that five out of 10 animals cannot enter into them. In how many ways will it be possible to accommodate ten animals in these ten cages?

Question

Dr. Abubacker · Accepted Answer

(b)
At first we have to a accommodate those 5 animals in cages which cannot enter in 4 small cages, therefore, number of ways are $\ ^{6}P_{5}$ and rest of the five animals arrange in 5! ways.
Total number of ways $= 5! 	imes^{6}P_{5}$
$= 120 	imes 720 = 86400$

Q. In a circus there are ten cages for accommodating ten animals. Out of these four cages are so small that five out of 10 animals cannot enter into them. In how many ways will it be possible to accommodate ten animals in these ten cages?

Explanation

Reviewed by Dr. Abubacker

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