# Math Help - How many ways can this type of 12 card hand be chosen?

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:

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

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

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:

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

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