How many decks of 13 cards are there that include at least one Jack, Queen, King or Ace?

I know you can subtract the amount of decks that don't include any of those, of which there are $\binom{36}{13}$, from the total number of possible decks, $\binom{52}{13}$, but I'd like to gain a better understanding of a different method:

can you solve it by designating a J/Q/K/A first (for which there are 16 choices) and then picking the other 12 cards? My initial thought was $16\binom{51}{12}$ but then I count decks multiple times. Say I designate the King of Hearts and then pick 12 random cards, one of which happens to be an Ace of Clubs. Later I designate the Ace of Clubs and pick 12 random cards, one of which happens to be the King of Hearts. This deck is counted double now. But how do I eliminate all multiple countings? How do I calculate the exact amount of multiple countings?

Thanks in advance.


Let $J$ denote the decks of $13$ cards out of $52$ that contain no Jack. Similar for $Q,K,A$.

Then with inclusion/exclusion and symmetry we find:$$|J\cup Q\cup K\cup A|=4|J|-6|J\cap Q|+4|J\cap Q\cap K|-|J\cap Q\cap K\cap A|=$$$$4\binom{48}{13}-6\binom{44}{13}+4\binom{40}{13}-\binom{36}{13}$$

So that $$|J^{\complement}\cap Q^{\complement}\cap K^{\complement}\cap A^{\complement}|=\binom{52}{13}-4\binom{48}{13}+6\binom{44}{13}-4\binom{40}{13}+\binom{36}{13}$$

  • $\begingroup$ This is a nice solution but not really an answer to my question. $\endgroup$ – Surzilla Aug 19 '18 at 10:58
  • 1
    $\begingroup$ Then do not accept it. But answering your question(s) point by point is quite cumbersome. Also the inclusion/exclusion method is exactly a way to eliminate double countings whatsoever. $\endgroup$ – drhab Aug 19 '18 at 11:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.