TITLE: Friends in a queue
QUESTION [2 upvotes]: We have a queue of N humans. At this queue there are two friends. What is the probability that between friends will be M humans. (M + 2 < N)
So, what we got:

total number of combinations  = N!;

Lets take some example:
N=10, M=3. For that case, between friends can be 3 humans, any humans, so its not just 3!, its something like (10-2)! but grouped by 3... and in that place I stuck.
Can someone help with direction to solving?

REPLY [0 votes]: Since only probability is asked for, we can just focus on ways of placing the "special" people ($A$ and $B$, say, with $m$ others in between)
For favorable placements, the $1st$ of them  can be in any position from $1$ to $(n-m-1)$,
so $A$ and $B$ can be placed in $2(n-m-1)$ ways against a total of $n(n-1)$ ways,
hence $Pr = \dfrac{2(n-m-1)}{n(n-1)}$
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