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.

• A. 66400
• B. 86400
• C. 96400
• D. None of these

Answer 1

The correct option is B 86400

At first we have to accommodate those 5 animals in cages which can not enter in 4 small cages,

$\therefore$Number of ways ${=}^6{P}_5$

Now after accommodating 5 animals we left with 5 cages and 5 animals,

$\therefore$ Number of ways $=5!$

Hence,

Required number of ways ${=}^6{P}_5\times 5!=\frac{6!}{(6-5)!}\times 5!=6!\times 5!=86400$