3
$\begingroup$

A special deck has 5 suits of 13 cards each, making a total of 65 cards.

We are dealing a 6-card hand from this deck. In how many ways one can pick a "well-suited" hand (defined below)?

A well-suited hand is one that contains at least one card of every suit, no more than one card of any kind (no pairs, etc.), and not all cards of consecutive kinds. An example of a well-suited hand is {2 of spades, 4 of hearts, 5 of diamonds, 8 of clubs, 10 of clubs, Jack of clovers}.

My attempt: Since there are 5 suits and we pick 6 cards, there will be one suit of which there will be 2 cards. We will pick one card from all other suits.

$$\binom{5}{1} \left (\binom{13}{2}-12 \right)11\cdot 10 \cdot 9 \cdot 8 $$

Explanation:

  • There are 12 ways to get consecutive numbers from the suit e.g. 1&2, 2&3 etc, hence subtracting 12 from $$\binom{13}{2}$$.
  • If we pick 2 & 4 from deck A, then we can't pick these numbers/ranks from any other deck (since pairs are not allowed), hence from the next deck, only 11 options remain, then 10 options remain, and so on.

I am not very sure if what I am doing is correct.

$\endgroup$
6
  • 1
    $\begingroup$ I agree with the start of all of this, the $\binom{5}{1}$, the $\binom{13}{2}$ and so on... I personally dislike seeing "choose 1" in a binomial coefficient... there is no reason to make it less readable. $\binom{5}{1}$ is simply $5$. $\binom{11}{1}$ is simply $11$. $\binom{n}{1}$ is simply $n$. Otherwise, the only thing wrong that I see here is the minus twelve. You seem to be interpreting the "not all cards of consecutive kinds" phrase incorrectly (or at least in a different way than I do). $\endgroup$
    – JMoravitz
    Commented Jul 12, 2023 at 14:20
  • 1
    $\begingroup$ That is to say, I would interpret $1\spadesuit 6\spadesuit 2\heartsuit 3\clubsuit 4\diamondsuit 5*$ to be a hand of cards of consecutive kinds (rearranged, the numbers $1,2,3,4,5,6$ are all consecutive). I would not consider $1\spadesuit 2\spadesuit 5\heartsuit 7\clubsuit 8\diamondsuit 10*$ to be a hand of cards of consecutive kinds (even though $1,2$ are consecutive, this fails to collectively be consecutive as there are gaps when sorted). You incorrectly interpret this as "the two cards from the same suit can't be consecutive" when I interpret the phrase to be talking about the hand. $\endgroup$
    – JMoravitz
    Commented Jul 12, 2023 at 14:24
  • $\begingroup$ @JMoravitz Point taken about the notation. Edited the post accordingly. Your interpretation of "not consecutive kinds" makes sense. How do I change the calculation? Do I just remove the 12? $\endgroup$
    – PGupta
    Commented Jul 12, 2023 at 14:31
  • $\begingroup$ I would first answer the question ignoring the "consecutive cards" clause. I would then answer the question where we specifically wanted consecutive cards and to avoid some of the frustration of overcounting, I might choose the ranks (numbers) of the cards before anything else, then choose the suits and how to divvy the selected cards up among the suits. $\endgroup$
    – JMoravitz
    Commented Jul 12, 2023 at 14:45
  • $\begingroup$ There are $~5~$ ways of selecting the repeated suit. Assume, without loss of generality that the repeated suit is spades. Then, there are $~\displaystyle \binom{6}{2} \times 4!~$ ways of distributing the 5 suits among the $~6~$ cards, with two of the cards being spades. Then, there are $~\displaystyle \left[ ~\binom{13}{6} - 9 ~\right] ~$ ways of selecting 6 ranks that are not all consecutive. This assumes that the 9 groups of consecutive ranks range from Ace-Six up to Nine-Ace. $\endgroup$
    – user2661923
    Commented Jul 12, 2023 at 14:58

1 Answer 1

Reset to default
2
$\begingroup$

To reiterate and collate comments:

Your initial thoughts were good but you seem to have interpreted the clause about consecutive ranks incorrectly to apply only to the two cards from the same suit rather than to the hand as a whole. That is to say, I interpret $1\spadesuit 6\spadesuit 2\heartsuit 3\clubsuit 4\diamondsuit 5*$ to be a hand of consecutive ranks.

To correct your calculation, I would first find the count where we don't care about that clause and then subtract from this the amount who violate that clause.

To find the amount where we don't care about that clause, it is simply the calculation you had but where we omit the $-12$.

To find the amount where we specifically want consecutive numbers, pick which run of six numbers they are. (It could be 1-6, or it could be 2-7, 3-8, ... 8-13)... (this is of course assuming that "Aces don't go both ways" and the typical meanings of "consecutive" as it relates to ranks that are considered "larger" than 10 in the case you used "Jacks" in place of 11's or "Kings" in place of 13, etc...). There are $8$ options (seen by looking at the rank of the smallest in the run). We then approach the same way as before by picking the suit that has two instances, then picking which two ranks (noting they are from those six in the run, not selected from the 13 overall) are used for that suit, then the rank used for the "smallest remaining suit", what rank is used for the next "smallest suit" and so on...

$$5\times \binom{13}{2}\times 11\times 10\times 9\times 8 - 8\times 5\times \binom{6}{2}\times 4\times 3\times 2\times 1$$

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .