To start with, this question has never been asked as how I am going to ask:
What is the probability that a five card poker hand will have two pairs (with no additional cards)?
Example of two-pair poker hand:
2${\heartsuit}$, 2${\spadesuit}$, 3${\clubsuit}$, 3${\diamondsuit}$, A${\spadesuit}$
We have to pick three types of cards from {2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, A}:
Number of ways of choosing three different numbers from set of 13 numbers = ${13 \choose 3}$
Number of ways of choosing suits for two of the pairs = ${4 \choose 2}^2$
Number of ways of choosing a suit for the side card = ${4 \choose 1}$
I believe the above way is correct and comprehensive way of counting all possible combinations. Am I wrong?
For context, the solution to one of the MIT questions isn't dividing by the extra 3:
Question 1 and Answer 1