Dealt foru cards, how many hands consit of 2 pairs? i have the answer

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Homework Help Overview

The discussion revolves around a combinatorial problem involving card hands, specifically how many four-card hands can consist of two pairs. The original poster expresses confusion regarding the counting method used to arrive at the answer of 2,808.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster questions the logic behind choosing two cards of the same denomination from a limited set of suits, seeking clarity on the combinatorial reasoning involved.

Discussion Status

Some participants have provided explanations regarding the combinatorial choices, clarifying the reasoning behind the calculations. The original poster appears to gain understanding from the responses, indicating a productive exchange of ideas.

Contextual Notes

The discussion includes references to standard deck constraints and the specific nature of card combinations, which may influence the interpretation of the problem.

mr_coffee
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Hello everyone. I was loooking over some notes and I noticed this problem i had a question mark next too.

Supose that you are dealt four cards instead of 5. How many 4 card hands consist of two pairs (two cards of one denomination and two cards of a second denomoniation)?

The answer is 2,808.

It was found by:
(13 choose 2) * (4 choose 2)^2

But I'm confused because if you are choosing 2 cards of one denomination, that's like choosing two 5's. There is only one 5 in every 13 cards or suites correct?

so if you have
(13 choose 2) that means your choosing 2 cards out of 1 suit weither it be hearts, diamonds, clubs or spades. So how can you be choosing 2 of the same numbers from a set of 13?

Or are they saying, they are chooosing 2 cards of 1 suit, such as a 4 and a 10, and later they are choosing 2 cards to match that suit, such as a 10H and a 10C, that would be one (4 choose 2) and the other would be 4H, and a 4D that would be the other (4 choose 2)?

I think i got it by writing this question but just making sure, thanks!
 
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Let nCr mean "n choose r".

It's 13C2 because we need to choose 2 numbers that will be the two doubles, and there are only 13 numbers to choose from. Hence, we choose 2 from the 13 numbers.

Now, for the first number we have chosen, we must choose 2 suits out of four. That's your standard deck, so we have 4C2. Similarly for the second card we have chosen for the other double, so we have 4C2 again.

Hence, (13C2)(4C2)(4C2) = (13C2)(4C2)^2.
 
Ahh I get it now, thanks again!
 
mr_coffee said:
Ahh I get it now, thanks again!

I'm glad you got it.

Now, I'm having a hard time with my Probability assignment myself. :frown:
 
Post it up! hah i doubt I could help though, at least i'll be ready for my stat classes coming up from all this probability in discrete math, wee.
 
Yeah, I should be ready for Stats next term though.

Not sure how far your Probability goes, but it does go into complicated territory.

I will be posting up questions soon. :smile:
 

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