# Combinatorics with cards

1. May 13, 2016

### davedave

1. The problem statement, all variables and given/known data
4 cards are dealt from a 52-card deck. How many hands contain 2 distinct pairs?

2. Relevant equations
This is an expression I come up with 13C1x4C2x12C1x4C2.

3. The attempt at a solution
This is how I approach it.
From the 13 ranks, I choose 1.
In this rank, I choose 2 cards from 4.

From the remaining 12 ranks, I choose 1.
In this rank, I choose 2 cards from 4.

Now, I do the following calculations with combinations.

13C1x4C2x12C1x4C2 = 5616.

My neighbor told me that it's wrong. He didn't explain why it was wrong. I cannot find any mistakes.
Please explain what I have done wrong. Thank you very much.

2. May 13, 2016

### Fightfish

Your method results in double-counting. Using $^{13}C_{1} \times ^{12}C_{1}$ counts the combination (x,y) twice - once as (x,y) and another time as (y,x), when these hands should considered as the same.

3. May 13, 2016

### Danielm

Assume that one hand is AA, JJ if the first rank is A and the second rank is J. A can't be counted in the second rank as you stated it.
Based on your counting, another way would be JJ,AA if the first rank is J. The A rank is still available for the second pair generating a double counting.

4. May 13, 2016

### Ray Vickson

You need to choose two distinct ranks from the 13 ranks; then for each chosen rank you need to pick 2 of the 4 cards in that rank.