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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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    $\begingroup$ Does "each color appear twice" mean "each suit appears twice"? If so, that would be $8$ cards. $\endgroup$
    – lulu
    Oct 2, 2018 at 19:49
  • $\begingroup$ @lulu Thank you, I have fixed it $\endgroup$
    – Aemilius
    Oct 2, 2018 at 19:51
  • $\begingroup$ I don't see any order dependence...For, say, the second case I first choose the two suits that will have two cards, then I pick the two cards for each of those suits, then I choose the one card for each of the other suits. Where does the order dependence come in? $\endgroup$
    – lulu
    Oct 2, 2018 at 19:53
  • $\begingroup$ What about this approach: Choose one card from each suit, and fill the rest $13 \cdot 13 \cdot 13 \cdot 13 \cdot \binom{48}{2}$ why would order matter here? $\endgroup$
    – Aemilius
    Oct 2, 2018 at 19:57
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    $\begingroup$ Well...that's just a bad approach. How would you distinguish between, say, A. choosing $A\spadesuit$ as part of the initial "one of each" and then choosing $2\spadesuit$ as one of the two strays and B. making the same choices in the opposite order? $\endgroup$
    – lulu
    Oct 2, 2018 at 19:59

1 Answer 1

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  • Case 1: There is one suit which has three occurrences and the remaining suits each have one occurrence

    • Pick which suit has the triple occurrence: $\binom{4}{1}$
    • Pick which three ranks appear for that suit: $\binom{13}{3}$
    • From those suits remaining, for the earliest remaining in alphabetical order pick which rank appears: $\binom{13}{1}$
    • From those suits remaining, for the earliest remaining in alphabetical order after having removed the previously used suit, pick which rank appears: $\binom{13}{1}$
    • From the final suit remaining, pick which rank appears: $\binom{13}{1}$

  • Case2: There are two suits, each of which have two occurrences and the remaining two suits each have one occurrence

    • $\vdots$

"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."

It counts $10$ of spades, $9$ of clubs and $8$ of diamonds as a different configuration than $9$ of spades, $8$ of clubs, $10$ of diamonds. It does not however make any reference to order in the hand that it occurs and it is unambiguous which cards are used if you follow the multiplication principle properly. The hand described with $A,K,Q$ of hearts, $10$ of spades, $9$ of clubs, $8$ of diamonds is counted exactly once and not more.

We were careful when phrasing the multiplication principle argument to refer to the suits in alphabetical order so as to have absolutely no ambiguity which suit/rank was chosen at each step and to avoid overcounting.

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