Question: A football team consists of 20 offensive and 20 defensive players. The players are to be paired in groups of 2 for the purpose of determining roommates. What is the probability that there are 2i offensive–defensive roommate pairs, i = 1,2,...,10?
20 Offensive (O), 20 Defensive (D) ==> 40 people total and 20 pairs
The question asks for the probability of 2i OD pairs. If we have 2i OD pairs, then we have (20−2i) OO and DD pairs. So I think the solution will be of form:
$\frac{(\mbox{total OD pair combinations}) \cdot (\mbox{total OO Comb)(total DD comb})}{\mbox{total unordered pair combinations} }$
Therefore, the answer will be:
$\cfrac{\binom{20}{2i}^2(2i)! \left[ \cfrac{(20-2i)!}{2^{10-i}(10-i)!} \right]^2} {\cfrac{40!}{2^{20}20! }} i=0,1,...,10$
My question is, when i=10, the numerator should be logically $(20!)$ But here the second part of the numerator becomes $\cfrac{0}{0}$. So I can't proceed further.
Please help me to solve this problem. Thanks in advance.