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