Need help with a certain method of solving counting problems (designating a fixed element and then choosing the rest).

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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 $binom3613$, from the total number of possible decks, $binom5213$, 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 $16binom5112$ 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.




combinatorics




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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 $binom3613$, from the total number of possible decks, $binom5213$, 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 $16binom5112$ 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.




    combinatorics




    share|cite|improve this question






















      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      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 $binom3613$, from the total number of possible decks, $binom5213$, 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 $16binom5112$ 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.




      combinatorics




      share|cite|improve this question












      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 $binom3613$, from the total number of possible decks, $binom5213$, 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 $16binom5112$ 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.





      combinatorics





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      asked Aug 19 at 10:31









      Surzilla

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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:$$|Jcup Qcup Kcup A|=4|J|-6|Jcap Q|+4|Jcap Qcap K|-|Jcap Qcap Kcap A|=$$$$4binom4813-6binom4413+4binom4013-binom3613$$



          So that $$|J^complementcap Q^complementcap K^complementcap A^complement|=binom5213-4binom4813+6binom4413-4binom4013+binom3613$$






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          • This is a nice solution but not really an answer to my question.
            – Surzilla
            Aug 19 at 10:58






          • 1




            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.
            – drhab
            Aug 19 at 11:05










          Your Answer




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          up vote
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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:$$|Jcup Qcup Kcup A|=4|J|-6|Jcap Q|+4|Jcap Qcap K|-|Jcap Qcap Kcap A|=$$$$4binom4813-6binom4413+4binom4013-binom3613$$



          So that $$|J^complementcap Q^complementcap K^complementcap A^complement|=binom5213-4binom4813+6binom4413-4binom4013+binom3613$$






          share|cite|improve this answer




















          • This is a nice solution but not really an answer to my question.
            – Surzilla
            Aug 19 at 10:58






          • 1




            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.
            – drhab
            Aug 19 at 11:05














          up vote
          2
          down vote













          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:$$|Jcup Qcup Kcup A|=4|J|-6|Jcap Q|+4|Jcap Qcap K|-|Jcap Qcap Kcap A|=$$$$4binom4813-6binom4413+4binom4013-binom3613$$



          So that $$|J^complementcap Q^complementcap K^complementcap A^complement|=binom5213-4binom4813+6binom4413-4binom4013+binom3613$$






          share|cite|improve this answer




















          • This is a nice solution but not really an answer to my question.
            – Surzilla
            Aug 19 at 10:58






          • 1




            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.
            – drhab
            Aug 19 at 11:05












          up vote
          2
          down vote










          up vote
          2
          down vote









          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:$$|Jcup Qcup Kcup A|=4|J|-6|Jcap Q|+4|Jcap Qcap K|-|Jcap Qcap Kcap A|=$$$$4binom4813-6binom4413+4binom4013-binom3613$$



          So that $$|J^complementcap Q^complementcap K^complementcap A^complement|=binom5213-4binom4813+6binom4413-4binom4013+binom3613$$






          share|cite|improve this answer












          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:$$|Jcup Qcup Kcup A|=4|J|-6|Jcap Q|+4|Jcap Qcap K|-|Jcap Qcap Kcap A|=$$$$4binom4813-6binom4413+4binom4013-binom3613$$



          So that $$|J^complementcap Q^complementcap K^complementcap A^complement|=binom5213-4binom4813+6binom4413-4binom4013+binom3613$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 19 at 10:49









          drhab

          87.7k541119




          87.7k541119











          • This is a nice solution but not really an answer to my question.
            – Surzilla
            Aug 19 at 10:58






          • 1




            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.
            – drhab
            Aug 19 at 11:05
















          • This is a nice solution but not really an answer to my question.
            – Surzilla
            Aug 19 at 10:58






          • 1




            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.
            – drhab
            Aug 19 at 11:05















          This is a nice solution but not really an answer to my question.
          – Surzilla
          Aug 19 at 10:58




          This is a nice solution but not really an answer to my question.
          – Surzilla
          Aug 19 at 10:58




          1




          1




          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.
          – drhab
          Aug 19 at 11:05




          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.
          – drhab
          Aug 19 at 11:05












           

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