Drawing tickets from urn problem

Question

There are m+n tickets in an urn which are numbered 1, 2,...., m+n. Suppose n tickets are drawn randomly from the urn. Then what is the probability that x of the tickets drawn will have numbers exceeding all numbers on the tickets left in the urn?

Answer 1

You can think of it this way: If $x$ of the tickets drawn will have numbers exceeding all numbers on the tickets left in the urn, this implies that the ticket $m+n-x$ will be in the urn, whilst tickets $m+n-(x-1)$ to tickets $m+n$ must be removed from the urn.

What this means is that, the number of ways to choose such a configuration is equivalent to that of choosing $n-x$ tickets from the bottom $m+n-x-1$ tickets. And since the total number of ways to draw $n$ tickets from $n+m$ tickets is ${n+m} \choose n$, we know that the probability that is desired is given by:

$$ \frac{{n+m-x-1} \choose {n-x}}{{n+m} \choose n} = \frac{{n+m-x-1} \choose {m-1}}{{n+m} \choose n}$$ Which stems from the fact that $$ {n \choose r} = {n \choose {n-r}} $$