First slide

Permutations

Question

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?

Moderate

A

66400
B

86400
C

96400
D

None of these

Solution

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
$= 5! \times^{6} P_{5} = 120 \times 720 = 86400$

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