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

• davedave
This can be done in ##^{13}C_{2}\times ^{4}C_{2}\times ^{4}C_{2}## ways. Therefore, the total number of hands containing 2 distinct pairs is ##13\times 12 \times 6 \times 6 = 5616##.In summary, the correct expression for calculating the number of hands containing 2 distinct pairs in a 52-card deck is ##^{13}C_{2}\times ^{4}C_{2}\times ^{4}C_{2}##. The mistake in the original attempt was double-counting combinations, resulting in an incorrect answer of 5616 instead of the correct answer of 5616.
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.

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.

davedave said:

## 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.

davedave said:

## 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.

## What is combinatorics with cards?

Combinatorics with cards is a branch of mathematics that deals with counting the possible arrangements of a deck of cards or a set of playing cards.

## How many ways can a deck of cards be shuffled?

A deck of cards can be shuffled in 52! (52 factorial) ways, which is approximately 8.0658 x 10^67.

## What is the probability of getting a flush in a 5-card poker hand?

The probability of getting a flush in a 5-card poker hand is approximately 0.00198, or about 0.2%.

## How many different poker hands are possible?

There are a total of 2,598,960 possible poker hands, as calculated by the combination formula nCr = n! / (r!(n-r)!), where n is the total number of cards (52) and r is the number of cards in a hand (5).

## What is the "birthday problem" in combinatorics with cards?

The "birthday problem" in combinatorics with cards refers to the probability of two people in a group having the same birthday when randomly selecting cards from a deck. This probability increases as the group size increases, and becomes more likely than not when the group size reaches 23.

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