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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.

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

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  • $\begingroup$ This is a nice solution but not really an answer to my question. $\endgroup$ – Surzilla Aug 19 '18 at 10:58
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    $\begingroup$ 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

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