Checking understanding using multiple ways to calculate a 2 pair in 5 card hand of poker

Question

In order to get a better understanding of basic probabilities, I have been trying to calculate poker hand probabilities in as many ways as I can think - for example calculating the numerator for a 2 pair hand in a 5 card hand could be done in the following ways:

1) Choose Any Card, pick 2 cards from the remaining 12 types that aren't the first card's type, for these 2 cards choose 2 out of 4 suits : ${52 \choose 1}\cdot{12 \choose 2}\cdot{4 \choose 2}^2$

2) Pick 2 cards from the 13 types, for these 2 cards choose 2 out of 4 suits, pick 1 card from the remaining cards left in the deck : ${13 \choose 2}\cdot{4 \choose 2}^2\cdot{52-8 \choose 1}$

3) Pick 1 cards from the 13 types, for this card choose 2 out of 4 suits, pick 1 card from the remaining 12 types, for this cards choose 2 out of 4 suits, divide by 2 as order doesn't matter and we will have over counted, pick 1 card from the remaining 11 type, choose 1 out of 4 suit : $\frac{{13 \choose 1}\cdot{4 \choose 2}\cdot{12 \choose 1}\cdot{4 \choose 2}}{2}{11 \choose 1}\cdot{4\choose1}$

What I would like to know is where my understanding is wrong in the following calculation:

Choose 3 cards from 13 types, for 2 of these pick 2 out of 4 suits, and for 1 pick 1 out of 4 suits : ${13\choose3}\cdot{4\choose2}^2\cdot{4\choose1}$

Here I am off by a factor of 3 which makes me believe I am thinking along the right lines, but I can't see why I would need to multiply this by 3.

Thanks in advance for your help.

Answer 1

You actually need to multiply by $3 = \frac{123552}{41184}$ rather than $2$

This $3$ represents the ${3\choose 2}$ ways of choosing the two of the three values which are pairs, or equivalently the ${3\choose 1}$ ways of choosing the one of the three values which is a singleton