The probability of being dealt a 2 pair 5 card hand... Given a standard playing card deck of 52 cards, what is the probability of being dealt a 2 pair 5 card hand consisting exactly of one pair of face cards and one pair of NOT face cards is?

First, I'll solve the problem assuming that it doesn't matter whether or not the fifth card matches either of the pairs.

There are $3 \cdot 4 = 12$ face cards in the deck, and $52 - 12 = 40$ non-face cards.
The number of choices for the positions of the pairs is
$$\binom{5}{2} \cdot \binom{3}{2}$$
The probability of getting the desired cards in a specific order is:
$$P_0 = \frac{12}{52} \cdot \frac{3}{51} \cdot \frac{40}{50} \cdot \frac{3}{49} \cdot \frac{48}{48}$$
$$P_0 = \frac{3}{13} \cdot \frac{1}{17} \cdot \frac{4}{5} \cdot \frac{3}{49}$$
$$P_0 = \frac{3 \cdot 2 \cdot 2 \cdot 3}{13 \cdot 17 \cdot 5 \cdot 7 \cdot 7}$$
$$P_0 = \frac{36}{54145}$$
The final probability is simply the product of this and the number of choices for positions of the pairs:
$$P = P_0 \cdot \binom{5}{2} \cdot \binom{3}{2} = \frac{216}{10829}$$

If it does matter that the fifth card does not match either of the pairs, then the last term in our calculation of $P_0$ becomes $44/48$ instead of $48/48$, which gives us a final answer instead of $$ P = \frac{198}{10829}$$
