### Problem 35 GRE0568

Posted:

**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

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