How many ways are there to choose from a deck of cards?

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The prompt is How many ways are there to choose 7 cards from a deck of 52 cards, given that one card from each suit must be chosen?



The way I solved this was by imagining we already took 1 card from each suit, now we are left with 3 cards to be picked from 4 suits, this can be done easily using stars and bars giving us $$4+3-1choose 3 = 20 space ways$$
The other way that I seemed to find on the internet that also makes sense goes like this.



  • Choosing 4 from 1 suit (4+1+1+1)
    $$4choose1cdot13choose4cdot13choose1^3$$


  • Choosing 2 from 3 suits (2+2+2+1)
    $$4choose1cdot13choose2^2cdot13choose1$$


  • Choosing 3 from 1 suit and 2 from 2 other (3+2+1+1)
    $$4choose1cdot13choose4cdot3choose1cdot13choose2cdot13choose1^2$$


Which also makes sense but gives obviously a different answer than the first method, what am I missing? Which way is the right way?







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  • 3




    First of all, your method does not specify the values of the three last cards to be chosen, only which suit they are. So you are off by a very large factor.
    – Servaes
    Aug 22 at 8:30







  • 5




    Second, by your method you count many ways twice or more. For example, when choosing exactly two hearts (say the king and queen), you could first choose the king when choosing one from each suit, and then the queen when picking the remaining three cards. But you could also first choose the queen, and then the king. The result is the same, but you will count both ways.
    – Servaes
    Aug 22 at 8:34















up vote
0
down vote

favorite
1












The prompt is How many ways are there to choose 7 cards from a deck of 52 cards, given that one card from each suit must be chosen?



The way I solved this was by imagining we already took 1 card from each suit, now we are left with 3 cards to be picked from 4 suits, this can be done easily using stars and bars giving us $$4+3-1choose 3 = 20 space ways$$
The other way that I seemed to find on the internet that also makes sense goes like this.



  • Choosing 4 from 1 suit (4+1+1+1)
    $$4choose1cdot13choose4cdot13choose1^3$$


  • Choosing 2 from 3 suits (2+2+2+1)
    $$4choose1cdot13choose2^2cdot13choose1$$


  • Choosing 3 from 1 suit and 2 from 2 other (3+2+1+1)
    $$4choose1cdot13choose4cdot3choose1cdot13choose2cdot13choose1^2$$


Which also makes sense but gives obviously a different answer than the first method, what am I missing? Which way is the right way?







share|cite|improve this question
















  • 3




    First of all, your method does not specify the values of the three last cards to be chosen, only which suit they are. So you are off by a very large factor.
    – Servaes
    Aug 22 at 8:30







  • 5




    Second, by your method you count many ways twice or more. For example, when choosing exactly two hearts (say the king and queen), you could first choose the king when choosing one from each suit, and then the queen when picking the remaining three cards. But you could also first choose the queen, and then the king. The result is the same, but you will count both ways.
    – Servaes
    Aug 22 at 8:34













up vote
0
down vote

favorite
1









up vote
0
down vote

favorite
1






1





The prompt is How many ways are there to choose 7 cards from a deck of 52 cards, given that one card from each suit must be chosen?



The way I solved this was by imagining we already took 1 card from each suit, now we are left with 3 cards to be picked from 4 suits, this can be done easily using stars and bars giving us $$4+3-1choose 3 = 20 space ways$$
The other way that I seemed to find on the internet that also makes sense goes like this.



  • Choosing 4 from 1 suit (4+1+1+1)
    $$4choose1cdot13choose4cdot13choose1^3$$


  • Choosing 2 from 3 suits (2+2+2+1)
    $$4choose1cdot13choose2^2cdot13choose1$$


  • Choosing 3 from 1 suit and 2 from 2 other (3+2+1+1)
    $$4choose1cdot13choose4cdot3choose1cdot13choose2cdot13choose1^2$$


Which also makes sense but gives obviously a different answer than the first method, what am I missing? Which way is the right way?







share|cite|improve this question












The prompt is How many ways are there to choose 7 cards from a deck of 52 cards, given that one card from each suit must be chosen?



The way I solved this was by imagining we already took 1 card from each suit, now we are left with 3 cards to be picked from 4 suits, this can be done easily using stars and bars giving us $$4+3-1choose 3 = 20 space ways$$
The other way that I seemed to find on the internet that also makes sense goes like this.



  • Choosing 4 from 1 suit (4+1+1+1)
    $$4choose1cdot13choose4cdot13choose1^3$$


  • Choosing 2 from 3 suits (2+2+2+1)
    $$4choose1cdot13choose2^2cdot13choose1$$


  • Choosing 3 from 1 suit and 2 from 2 other (3+2+1+1)
    $$4choose1cdot13choose4cdot3choose1cdot13choose2cdot13choose1^2$$


Which also makes sense but gives obviously a different answer than the first method, what am I missing? Which way is the right way?









share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Aug 22 at 8:25









Archetype2142

460313




460313







  • 3




    First of all, your method does not specify the values of the three last cards to be chosen, only which suit they are. So you are off by a very large factor.
    – Servaes
    Aug 22 at 8:30







  • 5




    Second, by your method you count many ways twice or more. For example, when choosing exactly two hearts (say the king and queen), you could first choose the king when choosing one from each suit, and then the queen when picking the remaining three cards. But you could also first choose the queen, and then the king. The result is the same, but you will count both ways.
    – Servaes
    Aug 22 at 8:34













  • 3




    First of all, your method does not specify the values of the three last cards to be chosen, only which suit they are. So you are off by a very large factor.
    – Servaes
    Aug 22 at 8:30







  • 5




    Second, by your method you count many ways twice or more. For example, when choosing exactly two hearts (say the king and queen), you could first choose the king when choosing one from each suit, and then the queen when picking the remaining three cards. But you could also first choose the queen, and then the king. The result is the same, but you will count both ways.
    – Servaes
    Aug 22 at 8:34








3




3




First of all, your method does not specify the values of the three last cards to be chosen, only which suit they are. So you are off by a very large factor.
– Servaes
Aug 22 at 8:30





First of all, your method does not specify the values of the three last cards to be chosen, only which suit they are. So you are off by a very large factor.
– Servaes
Aug 22 at 8:30





5




5




Second, by your method you count many ways twice or more. For example, when choosing exactly two hearts (say the king and queen), you could first choose the king when choosing one from each suit, and then the queen when picking the remaining three cards. But you could also first choose the queen, and then the king. The result is the same, but you will count both ways.
– Servaes
Aug 22 at 8:34





Second, by your method you count many ways twice or more. For example, when choosing exactly two hearts (say the king and queen), you could first choose the king when choosing one from each suit, and then the queen when picking the remaining three cards. But you could also first choose the queen, and then the king. The result is the same, but you will count both ways.
– Servaes
Aug 22 at 8:34











3 Answers
3






active

oldest

votes

















up vote
2
down vote



accepted











The way I solved this was by imagining we already took 1 card from each suit, now we are left with 3 cards to be picked from 4 suits, this can be done easily using stars and bars giving us




Stars-and-bars counts ways to place identical items into distinct containers.   Thus it is inapplicable to counting the ways to select cards, which are distinguishable by rank.   It is the wrong way to go.




The alternative method does consider the selection of rank.   This is the right thing to do.



$7$ can be partitioned into sums of four terms as $(4+1+1+1)$, $(3+2+1+1)$, and $(2+2+2+1)$.   We count ways to select suits for each amount, and ways to select those amounts of ranks in the selected suits.



$$binom 4 1,3binom 134^1binom131^3+binom41,1,2binom133^1binom132^1binom131^2+binom 43,1binom132^3binom131^1$$






share|cite|improve this answer



























    up vote
    2
    down vote













    This does not answer your question, but provides a route and is too large for a comment.




    Let $S$ denote the collection of sets of $7$ cards that contain at least one spade. Similarly for $D,H,C$.



    The collection of sets of $7$ cards that contain no spades is denoted by $S^complement$.



    Then by inclusion/exclusion and symmetry:



    $$|S^complementcup D^complementcup H^complementcup C^complement|=binom 41|S^complement|-binom42|S^complementcap D^complement|+binom43|S^complementcap D^complementcap H^complement|$$leading to:$$|Scap Dcap Hcap C|=binom527-binom41binom397+binom42binom267-binom43binom137$$






    share|cite|improve this answer



























      up vote
      0
      down vote













      So we have two ways to solve this question:
      $1.$Find the number of ways that we can choose 7 cards that does not include all the 4 suits and then subtract the answer from the number of all the possible ways that we can choose 7 cards.
      $2.$Find the number of ways that we can choose 7 cards from the deck with all the suits included.
      The second one is easier so:
      The number of the cards from each suit can be:
      $4/1/1/1$ ,
      $3/2/1/1$ or
      $2/2/2/1$
      I’m going to use the same approach as your second solution but please note that your second solution is also wrong because for example , in $2/2/2/1$ , it should be $P(13;11,2) ^ 3$ not $P(13;11,2) ^ 2$.
      I’m not going to give you the answer but please note that your second approach is right you just made a mistake in the solution.
      And the reason that your first solution is wrong is because you are just calculating the number of ways you can pick 3 cards from 6 cards which is obviously wrong since , let’s suppose we are using the $4/1/1/1$ combination(Let’s suppose we have 4 spades) , the number of ways you can pick 4 spades is alone bigger than your answer.
      Hope this helps!






      share|cite|improve this answer




















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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        2
        down vote



        accepted











        The way I solved this was by imagining we already took 1 card from each suit, now we are left with 3 cards to be picked from 4 suits, this can be done easily using stars and bars giving us




        Stars-and-bars counts ways to place identical items into distinct containers.   Thus it is inapplicable to counting the ways to select cards, which are distinguishable by rank.   It is the wrong way to go.




        The alternative method does consider the selection of rank.   This is the right thing to do.



        $7$ can be partitioned into sums of four terms as $(4+1+1+1)$, $(3+2+1+1)$, and $(2+2+2+1)$.   We count ways to select suits for each amount, and ways to select those amounts of ranks in the selected suits.



        $$binom 4 1,3binom 134^1binom131^3+binom41,1,2binom133^1binom132^1binom131^2+binom 43,1binom132^3binom131^1$$






        share|cite|improve this answer
























          up vote
          2
          down vote



          accepted











          The way I solved this was by imagining we already took 1 card from each suit, now we are left with 3 cards to be picked from 4 suits, this can be done easily using stars and bars giving us




          Stars-and-bars counts ways to place identical items into distinct containers.   Thus it is inapplicable to counting the ways to select cards, which are distinguishable by rank.   It is the wrong way to go.




          The alternative method does consider the selection of rank.   This is the right thing to do.



          $7$ can be partitioned into sums of four terms as $(4+1+1+1)$, $(3+2+1+1)$, and $(2+2+2+1)$.   We count ways to select suits for each amount, and ways to select those amounts of ranks in the selected suits.



          $$binom 4 1,3binom 134^1binom131^3+binom41,1,2binom133^1binom132^1binom131^2+binom 43,1binom132^3binom131^1$$






          share|cite|improve this answer






















            up vote
            2
            down vote



            accepted







            up vote
            2
            down vote



            accepted







            The way I solved this was by imagining we already took 1 card from each suit, now we are left with 3 cards to be picked from 4 suits, this can be done easily using stars and bars giving us




            Stars-and-bars counts ways to place identical items into distinct containers.   Thus it is inapplicable to counting the ways to select cards, which are distinguishable by rank.   It is the wrong way to go.




            The alternative method does consider the selection of rank.   This is the right thing to do.



            $7$ can be partitioned into sums of four terms as $(4+1+1+1)$, $(3+2+1+1)$, and $(2+2+2+1)$.   We count ways to select suits for each amount, and ways to select those amounts of ranks in the selected suits.



            $$binom 4 1,3binom 134^1binom131^3+binom41,1,2binom133^1binom132^1binom131^2+binom 43,1binom132^3binom131^1$$






            share|cite|improve this answer













            The way I solved this was by imagining we already took 1 card from each suit, now we are left with 3 cards to be picked from 4 suits, this can be done easily using stars and bars giving us




            Stars-and-bars counts ways to place identical items into distinct containers.   Thus it is inapplicable to counting the ways to select cards, which are distinguishable by rank.   It is the wrong way to go.




            The alternative method does consider the selection of rank.   This is the right thing to do.



            $7$ can be partitioned into sums of four terms as $(4+1+1+1)$, $(3+2+1+1)$, and $(2+2+2+1)$.   We count ways to select suits for each amount, and ways to select those amounts of ranks in the selected suits.



            $$binom 4 1,3binom 134^1binom131^3+binom41,1,2binom133^1binom132^1binom131^2+binom 43,1binom132^3binom131^1$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Aug 23 at 0:05









            Graham Kemp

            80.6k43275




            80.6k43275




















                up vote
                2
                down vote













                This does not answer your question, but provides a route and is too large for a comment.




                Let $S$ denote the collection of sets of $7$ cards that contain at least one spade. Similarly for $D,H,C$.



                The collection of sets of $7$ cards that contain no spades is denoted by $S^complement$.



                Then by inclusion/exclusion and symmetry:



                $$|S^complementcup D^complementcup H^complementcup C^complement|=binom 41|S^complement|-binom42|S^complementcap D^complement|+binom43|S^complementcap D^complementcap H^complement|$$leading to:$$|Scap Dcap Hcap C|=binom527-binom41binom397+binom42binom267-binom43binom137$$






                share|cite|improve this answer
























                  up vote
                  2
                  down vote













                  This does not answer your question, but provides a route and is too large for a comment.




                  Let $S$ denote the collection of sets of $7$ cards that contain at least one spade. Similarly for $D,H,C$.



                  The collection of sets of $7$ cards that contain no spades is denoted by $S^complement$.



                  Then by inclusion/exclusion and symmetry:



                  $$|S^complementcup D^complementcup H^complementcup C^complement|=binom 41|S^complement|-binom42|S^complementcap D^complement|+binom43|S^complementcap D^complementcap H^complement|$$leading to:$$|Scap Dcap Hcap C|=binom527-binom41binom397+binom42binom267-binom43binom137$$






                  share|cite|improve this answer






















                    up vote
                    2
                    down vote










                    up vote
                    2
                    down vote









                    This does not answer your question, but provides a route and is too large for a comment.




                    Let $S$ denote the collection of sets of $7$ cards that contain at least one spade. Similarly for $D,H,C$.



                    The collection of sets of $7$ cards that contain no spades is denoted by $S^complement$.



                    Then by inclusion/exclusion and symmetry:



                    $$|S^complementcup D^complementcup H^complementcup C^complement|=binom 41|S^complement|-binom42|S^complementcap D^complement|+binom43|S^complementcap D^complementcap H^complement|$$leading to:$$|Scap Dcap Hcap C|=binom527-binom41binom397+binom42binom267-binom43binom137$$






                    share|cite|improve this answer












                    This does not answer your question, but provides a route and is too large for a comment.




                    Let $S$ denote the collection of sets of $7$ cards that contain at least one spade. Similarly for $D,H,C$.



                    The collection of sets of $7$ cards that contain no spades is denoted by $S^complement$.



                    Then by inclusion/exclusion and symmetry:



                    $$|S^complementcup D^complementcup H^complementcup C^complement|=binom 41|S^complement|-binom42|S^complementcap D^complement|+binom43|S^complementcap D^complementcap H^complement|$$leading to:$$|Scap Dcap Hcap C|=binom527-binom41binom397+binom42binom267-binom43binom137$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Aug 22 at 9:12









                    drhab

                    88k541120




                    88k541120




















                        up vote
                        0
                        down vote













                        So we have two ways to solve this question:
                        $1.$Find the number of ways that we can choose 7 cards that does not include all the 4 suits and then subtract the answer from the number of all the possible ways that we can choose 7 cards.
                        $2.$Find the number of ways that we can choose 7 cards from the deck with all the suits included.
                        The second one is easier so:
                        The number of the cards from each suit can be:
                        $4/1/1/1$ ,
                        $3/2/1/1$ or
                        $2/2/2/1$
                        I’m going to use the same approach as your second solution but please note that your second solution is also wrong because for example , in $2/2/2/1$ , it should be $P(13;11,2) ^ 3$ not $P(13;11,2) ^ 2$.
                        I’m not going to give you the answer but please note that your second approach is right you just made a mistake in the solution.
                        And the reason that your first solution is wrong is because you are just calculating the number of ways you can pick 3 cards from 6 cards which is obviously wrong since , let’s suppose we are using the $4/1/1/1$ combination(Let’s suppose we have 4 spades) , the number of ways you can pick 4 spades is alone bigger than your answer.
                        Hope this helps!






                        share|cite|improve this answer
























                          up vote
                          0
                          down vote













                          So we have two ways to solve this question:
                          $1.$Find the number of ways that we can choose 7 cards that does not include all the 4 suits and then subtract the answer from the number of all the possible ways that we can choose 7 cards.
                          $2.$Find the number of ways that we can choose 7 cards from the deck with all the suits included.
                          The second one is easier so:
                          The number of the cards from each suit can be:
                          $4/1/1/1$ ,
                          $3/2/1/1$ or
                          $2/2/2/1$
                          I’m going to use the same approach as your second solution but please note that your second solution is also wrong because for example , in $2/2/2/1$ , it should be $P(13;11,2) ^ 3$ not $P(13;11,2) ^ 2$.
                          I’m not going to give you the answer but please note that your second approach is right you just made a mistake in the solution.
                          And the reason that your first solution is wrong is because you are just calculating the number of ways you can pick 3 cards from 6 cards which is obviously wrong since , let’s suppose we are using the $4/1/1/1$ combination(Let’s suppose we have 4 spades) , the number of ways you can pick 4 spades is alone bigger than your answer.
                          Hope this helps!






                          share|cite|improve this answer






















                            up vote
                            0
                            down vote










                            up vote
                            0
                            down vote









                            So we have two ways to solve this question:
                            $1.$Find the number of ways that we can choose 7 cards that does not include all the 4 suits and then subtract the answer from the number of all the possible ways that we can choose 7 cards.
                            $2.$Find the number of ways that we can choose 7 cards from the deck with all the suits included.
                            The second one is easier so:
                            The number of the cards from each suit can be:
                            $4/1/1/1$ ,
                            $3/2/1/1$ or
                            $2/2/2/1$
                            I’m going to use the same approach as your second solution but please note that your second solution is also wrong because for example , in $2/2/2/1$ , it should be $P(13;11,2) ^ 3$ not $P(13;11,2) ^ 2$.
                            I’m not going to give you the answer but please note that your second approach is right you just made a mistake in the solution.
                            And the reason that your first solution is wrong is because you are just calculating the number of ways you can pick 3 cards from 6 cards which is obviously wrong since , let’s suppose we are using the $4/1/1/1$ combination(Let’s suppose we have 4 spades) , the number of ways you can pick 4 spades is alone bigger than your answer.
                            Hope this helps!






                            share|cite|improve this answer












                            So we have two ways to solve this question:
                            $1.$Find the number of ways that we can choose 7 cards that does not include all the 4 suits and then subtract the answer from the number of all the possible ways that we can choose 7 cards.
                            $2.$Find the number of ways that we can choose 7 cards from the deck with all the suits included.
                            The second one is easier so:
                            The number of the cards from each suit can be:
                            $4/1/1/1$ ,
                            $3/2/1/1$ or
                            $2/2/2/1$
                            I’m going to use the same approach as your second solution but please note that your second solution is also wrong because for example , in $2/2/2/1$ , it should be $P(13;11,2) ^ 3$ not $P(13;11,2) ^ 2$.
                            I’m not going to give you the answer but please note that your second approach is right you just made a mistake in the solution.
                            And the reason that your first solution is wrong is because you are just calculating the number of ways you can pick 3 cards from 6 cards which is obviously wrong since , let’s suppose we are using the $4/1/1/1$ combination(Let’s suppose we have 4 spades) , the number of ways you can pick 4 spades is alone bigger than your answer.
                            Hope this helps!







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Aug 22 at 18:28









                            Borna Ahmadzade

                            297




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