In how many ways can one pick 6 cards from a deck of 52 cards so that all suits are present?
This problem is troubling me for a while. The expected answer to the problem is: $$\binom{4}{1}\binom{13}{3}\binom{13}{1} \binom{13}{1} \binom{13}{1} + \binom{4}{2}\binom{13}{2}\binom{13}{2} \binom{13}{1} \binom{13}{1} $$
This is how I understand it:
There are two possible scenarios:
- One suit appears three times, the others only once
- Two suits appear twice, and two appear once.
However, I am of the mind that the solution is not correct - it seems to be order-dependent. Let's say that Hearts will appear three times - we draw A, K and Q of hearts - here, order does not matter.
But now, let's say that we also wanna take 10 of spades, 9 of clubs and 8 of diamonds.
Here, it seems to me that this solution counts
(10)(9)(8), (9)(8)(10), (8)(10)(9), etc.. as different configurations.
Could you, please, explain to me where my reasoning went wrong?