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Question

Mathematics Question on permutations and combinations

A teaparty is arranged for 16 people along two sides of a large table with 8 chairs on each side. Four men want to sit on one particular side and two on the other side. The number of ways in which they can be seated is

A

6!8!10!4!6!\frac{6!\,8!\,10!}{4!\,6!}

B

8!8!10!4!6!\frac{8!\,8!\,10!}{4!\,6!}

C

8!8!6!6!4!\frac{8!\,8!\,6!}{6!\,4!}

D

None of these

Answer

8!8!10!4!6!\frac{8!\,8!\,10!}{4!\,6!}

Explanation

Solution

There are 8 chairs on each side of the table. Let the sides be represented by A and B. Let four persons sit on side A, then number of ways of arranging 4 persons on 8 chairs on side A=8P4A = ^8P_4 and then two persons sit on side B. The number of ways of arranging 2 persons on 8 chairs on side B=8P2B = ^8P_2 and the remaining 10 persons can be arranged in remaining 10 chairs in 10! ways. Hence the total number of ways in which the persons can be arranged =8P4×8P2×10!=8!8!10!4!6!=^{8}P_{4}\times^{8}P_{2}\times10!=\frac{8!\,8!\,10!}{4!\,6!}