Thread:

1. How many ways can this type of 12 card hand be chosen?

How many ways can 12 cards be chosen from a standard deck of 52 cards such that the same number of cards from each suit are in the 12?

The way I saw it was that initially, you had to chose 1 of the 4 suits, i.e., $4\choose{1}$. Then within the chosen suit you needed to choose 3 cards, i.e., $13\choose{3}$, then repeat the process with the remaining 3 suits, so my solution was:

$4\choose{1}$ $13 \choose{3}$ $3\choose{1 }$ $13\choose{3}$ $2\choose{1}$ $13\choose{3}$ $1\choose{1}$ $13 \choose{3}$

The answer provided says:

$13 \choose{3}$ $13 \choose{3}$ $13 \choose{3}$ $13 \choose{3}$

Why is the choice of suit not included?

2. Re: How many ways can this type of 12 card hand be chosen?

Originally Posted by terrorsquid
How many ways can 12 cards be chosen from a standard deck of 52 cards such that the same number of cards from each suit are in the 12?

The way I saw it was that initially, you had to chose 1 of the 4 suits, i.e., $4\choose{1}$. Then within the chosen suit you needed to choose 3 cards, i.e., $13\choose{3}$, then repeat the process with the remaining 3 suits, so my solution was:

$4\choose{1}$ $13 \choose{3}$ $3\choose{1 }$ $13\choose{3}$ $2\choose{1}$ $13\choose{3}$ $1\choose{1}$ $13 \choose{3}$

The answer provided says:

$13 \choose{3}$ $13 \choose{3}$ $13 \choose{3}$ $13 \choose{3}$

Why is the choice of suit not included?
13 choose 3 indicates you are picking one suit. You are doing that 4 times i.e. all 4 suits are being considered.

3. Re: How many ways can this type of 12 card hand be chosen?

I realize this is a few days old and dwsmith covered it quite succinctly. However, just in case: You are mixing up probability with choices. There is no probability in this question at all. Its 100% likely you are going to get 12 cards, in which each suit has 3 cards. Imagine the whole deck is laid out in front of you face up. You can pick any 3 hearts you want, any 3 spades you want ,etc.

How many different ways can you do that? dwsmith has the answer right above.
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