Given a standard playing card deck of 52 cards. The probability of being dealt a 2 pair 5 card hand consisting exactly two pair of NOT face cards Homework. A face card is a king, Queen or Jack. Four of a kind is not considered two pair
 A: Two pairs is the trickiest of the standard "poker hand" problems, the easiest one to get wrong. As usual we note that there are $\binom{52}{5}$ possible hands, all equally likely. We need to count the "favourables."
The two kinds we have $2$ each of can be chosen in $\binom{10}{2}$ ways. For there are $3$ kinds of face cards, and hence $10$ kinds of non-face cards. 
For each of the $2$ kinds we have chosen, there are $\binom{4}{2}$ ways to choose the actual cards of the higher-ranking kind, and for each of these there are $\binom{4}{2}$ ways to choose the actual cards of the lower-ranking kind. 
For each way of doing these things, there are $\binom{44}{1}$ ways to choose the useless fifth card, since we must avoid the $2$ kinds we have pairs in.
Thus the number of favourables is $\binom{10}{2}\binom{4}{2}\binom{4}{2}\binom{44}{1}$.
Remark: It is all too easy to decide that there are (in this problem) $(10)(9)$ ways to choose the kinds. That would be double-counting, since two $7$'s and two Aces is the same as two Aces and two $7$'s. But we could deliberately double-count, and then divide by $2$. 
