Problem 35 GRE0568

Forum for the GRE subject test in mathematics.
EmLasker
Posts: 24
Joined: Sun Oct 05, 2008 1:41 am

Problem 35 GRE0568

Postby EmLasker » Sun Oct 12, 2008 11:43 am

The question is: At a banquet, 9 women and 6 men are to be seated in a row of 15 chairs. If the entire seating arrangement is chosen at random, what is the probability that all 6 men will sit next to each other in 6 consecutive positions?

My attempt at a solution:
First, I noted that there were 10 different groups of seats for the six men to sit in (starting with seats 1-6, then 2-7 up to seats 10-15). There are 6! permutations of the six men and 15! possible seating arrangements. Thus I am getting 10*6! possible ways for the 6 men to sit together, so that the probability is 10*(6!)/15!

However, the answer is (10!)*(6!)/15!

I just don't see where the extra factorial is coming from - I looked up some similar problems and it always seems to be there...


Peter

amateur
Posts: 42
Joined: Wed Sep 10, 2008 9:41 am

Postby amateur » Sun Oct 12, 2008 11:50 am

Hi,

I did it the following way.

Consider the six men (call them A, B, C, D, E, F) as a single group.
Next consider the nine women each as a separate element (call them 1, 2, 3, 4, 5, 6, 7, 8, 9).

There are a total of 10 elements to be permuted (1, 2, 3, 4, 5, 6, 7, 8, 9, ABCDEF). This can be done in 10! ways.

Further, the ABCDEF can be permuted in 6! ways. Hence the total# of ways to seat the six men consecutively is 10! 6!.

Divide this by all possible (15!) permutations and you get the answer 10! 6! / 15!

EmLasker
Posts: 24
Joined: Sun Oct 05, 2008 1:41 am

Postby EmLasker » Sun Oct 12, 2008 12:01 pm

Amateur,

Thank you so much - I see it now. I think my solution only counted the possible ways for the men to sit consecutively, without counting the various permutations of the women as a separate seating arrangement.

Peter




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