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., $\displaystyle 4\choose{1}$. Then within the chosen suit you needed to choose 3 cards, i.e., $\displaystyle 13\choose{3}$, then repeat the process with the remaining 3 suits, so my solution was:

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

The answer provided says:

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

Why is the choice of suit not included?