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Permutation Combination Problem 2

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by ladhanivishal » Tue Jul 07, 2009 2:32 am
At a dinner party, 5 people are to be seated around a circular table. Two seating arrangements are considered different only when the positions of the people are different relative to each other. What is the total nummber of different possible seating arrangements for the group?

(A) 5
(B) 10
(C) 24
(D) 32
(E) 120

Please let me know how to go about solving this?
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Source: — Problem Solving |

by GMATQuantCoach » Tue Jul 07, 2009 4:43 am
First pretend that they are not in a circle. There are 5! ways to arrange them.

I'm going to name the 5 people as A,B,C,D, and E.

Examples of permutations of them when not in a circle are:

ABCDE
EABCD
DEABC
CDEAB
BCDEA

If they were in a circle, the 5 permutations above count as 1. We can rotate them, and still the permutations are considered to be the same. Hence we need to discount the number of permutations by 5.

5!/5 = 24

Answer is C.
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