Combinatorics with cards

  • Thread starter davedave
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  • #1
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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.
 

Answers and Replies

  • #2
954
117
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
21
0

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.
 
  • #4
Ray Vickson
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Homework Helper
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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.
 

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