How Many Ways to Deal Two Distinct Pairs in a Four-Card Hand?

[davedave]

Homework Statement

4 cards are dealt from a 52-card deck. How many hands contain 2 distinct pairs?

Homework Equations

This is an expression I come up with 13C1x4C2x12C1x4C2.

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.

 

[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.

 

[Danielm]

Homework Statement

4 cards are dealt from a 52-card deck. How many hands contain 2 distinct pairs?

Homework Equations

This is an expression I come up with 13C1x4C2x12C1x4C2.

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.

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.

 

[Ray Vickson]

Homework Statement

4 cards are dealt from a 52-card deck. How many hands contain 2 distinct pairs?

Homework Equations

This is an expression I come up with 13C1x4C2x12C1x4C2.

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.

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.