# Showing that the number of queues is uniquely expresssed as the product of $2010$ positive integers.

This problem came from a friend in preparation for a contest.

There are $$2011$$ people in a queue lining up for a conference, and no two people have the same height. Bob is the $$27$$th tallest person in the group. Let $$n$$ be the number of queues that the group of $$2011$$ people can form such that in the queue, Bob is shorter than everyone else in front of him.

Prove that there are exactly one set $$S$$ of $$2010$$ distinct positive integers, such that $$n$$ is the product of all the elements of $$S$$.

My work:

Number the queue $$1,2,\dots, 2011$$. Let $$b$$ be Bob's position. Clearly, $$b\leq 27$$ since if $$b=28$$, there is guaranteed to be a person in front of him that's shorter.

If $$b=27$$, there are $$26$$ spaces in front of him, and $$2011-27=1984$$ places behind him. So the number of queues in this case is $$26!\times 1984!$$.

if $$b=26$$, we obtain similarly $$25!\times 1985!$$. Continuing down to $$b=1$$, we find that the total number of queues is $$n=\sum_{i=1}^{26}i!\times (2010-i)!$$ From here, I thought we could find the elements of $$S$$ by factoring something like $$n=1984!\times (\dots)$$ but I'm not quite sure how to continue.

There are $$2011!$$ possible queues without restriction, but we count only those where Bob appears as first among the $$27$$ tallest people. By symmetry, this happens in precisely $$\frac1{27}$$ of all queues, so $$n=\frac{2011!}{27}$$ and this immediately suggests one possible way to write $$n$$ as product of $$2010$$ distinct positive integers, namely the integers $$1,2,\ldots,2011$$ except $$27$$.
Any other product of $$2010$$ distinct positive integers is either of the form $$\frac{2011!}{k}$$ with $$1\le k\le 2011$$ (and obviously no other $$k$$ tan $$27$$ will give the correct value), or has at least one factor $$\ge 2012$$ and the product of the remaining $$2009$$ distinct positive integers is $$\ge 2009!$$, so in total the result would be too big.