Below given is a question on Combinatorics :
$2n$ players participate in the first round of a tennis tournament: $k$ females and $2n - k$ males. n single matches are fixed at random by successively drawing, without replacement, the names of all $2n$ players from an urn: the player drawn first plays against the one whose name comes up second, the third against the fourth, etc.
What is the probability that there will not be a single match involving two female players?
Answer : I tried it using the following approach, but my answer turns out to be wrong, though I can't find a mistake in the approach I followed. Can someone please help me figure out my mistake?
The total number of ways in which these $2n$ players can be arranged for n matches is $(2n)!$ .
For counting the favourable ways ( number of arrangements of these n players such that no $2$ females play each other ) I first arrange the $2n - k$ male players in a line in $(2n-k)!$ ways. Now this arrangement creates $2n-k+1$ empty spaces for the females to occupy( +1 because the position after the last male player can also be occupied). Of these $2n-k+1$ positions if we select any k of them for the females, then the arrangement formed will guarantee that no $2$ females play each other.
Hence the number of favourable ways to arrange the 2n players is $(2n-k)! *k! * {2n-k+1\choose k}$.
Thus the required probability is $\frac{2n-k+1\choose k}{2n\choose k} $
And the correct answer is $2^k * \frac{n\choose k}{2n\choose k}$.
You are undercounting. Your probability will always be lesser than the actual answer. So, here is one of the cases you are missing: You are assuming that no girls can be together in any of the valid permutation, which is a subset of all the valid permutations. Assume k = 2 and n = 3. Then a valid permutation is |BG|GB|BB| (vertical lines for better visualisation of teams being formed). But, this will never be counted in your method.