Permutations And Combinations Question 284

In a circus there are ten cages for accommodating ten animals. Out of these, four cages are so small that five out of the ten animals cannot enter into them. In how many ways will it be possible to accommodate ten animals in these ten cages [Roorkee 1989]

Options:

A) 66400

B) 86400

C) 96400

D) None of these

Show Answer

Answer:

Correct Answer: B

Solution:

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



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