How many ways are there to choose 6 cards from a standard 52-card deck?

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?

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