Problem. A bridge player and his partner are known to have six spades between them. Find the the probability that the spades will be split 3-3.
Solution.
In the 26-26 distribution of cards between the two pairs of partners we want that in one pair the partners receive 3 spades each and the second receives the remaining spades at any pattern. Thus we only have to deal with the distribution of the 6 spades within the first pair.
In particular, it suffices to fill the hand of one of the partners, since the other is automatically determined. For one of the partners, choose 3 spades out of 6 in ${6\choose 3}$ ways, then choose 10 cards from the remaining 20. So there are $$ {6\choose 3}{20\choose10} $$ overall ways to distribute the 6 spades among the partners s.t. they both have 3 spades.
Furthermore, there are ${26 \choose 13}$ overall ways to draw 26 cards for two partners, so the probability is $$ \frac{{6\choose 3}{20\choose10}}{{26 \choose 13}}.\quad(*) $$
Now the question is: if the pair is not fixed beforehand, is the answer equal to $2!$ times $(*)$?