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Probablity - Permutations with Alernate Seating

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Probablity - Permutations with Alernate Seating

by skalevar » Mon Sep 13, 2010 5:35 am
From Veritas Test Prep Cominatorics & Probablity p. 60

In how many ways can 3 men and 3 women be seated in 6 seats if they must alternate?

My answer:
There are 3 pairs of man/woman. Therefor, there are 3! ways of arranging these pairs, but each of these pairs could be arrange in two ways (M-W or W-M). Therefore, I multiply 3! x 2x2x2 = 3!x2^3 = 48

This is incorrect. Why?

Veritas Answer:
72
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by alivapriyada » Mon Sep 13, 2010 6:31 am
The places can be filled up starting with a man or a woman
and for each possibility it can be filled up in the way stated below.
_ _ _ _ _ _
3*3*2*2*1*1=36

so total no of ways=36*2

hope it helps.
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by Maciek » Mon Sep 13, 2010 6:34 am
Hi!

This question is tricky. I hope I understand it correctly.

They must alternate, so we have two possibilities:
- Men take even seats and women take odd seats
- Women take even seats and men take odd seats

3 men can take 3 seats. Therefore, they can be seated in 3! ways.
3 women can take 3 seats. Therefore, they can be seated in 3! ways.

Hence, 3 men and 3 women can be seated in 6 seats in x ways, if they must alternate.
x = 2*3!*3! = 72

Hope it helps!
Best,
Maciek
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