At a banquet, 9 women and 6 men are to be seated in a row of 15 chairs. If the entire seating arrangement is to be chosen at random, what is the probability that all of the men will be seated next to each other in 6 consecutive positions?
The correct answer is $\displaystyle \frac{6! 10!}{15!}$. Why is it not $\displaystyle \frac{6! 9!}{15!}$?
It seems to me it should be, because, there are $15!$ total possible ways to order 15 people without regard to order. Then, $6!$ ways of filling the first 6 chairs, and $9!$ ways of filling the rest. So, it should be $6! 9!$ divided by $15!$. Where does the $10!$ come from, and what am I missing here?
Best Answer
Another way to view this problem is:
Put the $6$ men in order ($6!$ ways).
Now consider the $6$ ordered men as a single unit.
Arrange the $10$ items ($9$ women and one block of $6$ ordered men) in some order. ($10!$ ways).