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 is possible to accommodate ten animals in these cages

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

number of ways are ${}^{6}P_5$. Now after accommodating 5 animals we left with 5 cages and 5 animals, therefore

$\text{ number of ways are }5!\text{ . Hence required number of ways }$$={}^{6}P_5\times 5!=86400$