Four married couples attend a party. Each person shakes hands with every other person, except their own...












1












$begingroup$


My book gave the answer as 24. I thought of it like this:



You have four pairs of couples, so you can think of it as M1W2, M2W2, M3W3, M4W4, where
M is a man and W is a woman. M1 has to shake 6 other hands, excluding his wife. You have to do this 4 times for the other men, so you have 4 * 6 handshakes, but in my answer you are double counting. How do I approach this?






combinatorics







share|cite|improve this question









$endgroup$












  • $begingroup$
    In your answer, you both overcounted and undercounted, and incidentally these happened to cancel out and give you the correct answer without having to do anything further. You did $4 times (text{Handshakes done by the men})$, which overcounted the man-man handshakes, but left out the woman-woman handshakes.
    $endgroup$
    – M. Vinay
    38 mins ago










  • $begingroup$
    And that's easily fixed by counting all such handshakes in the same way, not just those done by men, so you get $48$. And now, as you said, you have indeed double-counted. But if you know it's exactly double counting, you can get the answer by halving it!
    $endgroup$
    – M. Vinay
    30 mins ago










  • $begingroup$
    I recommend when you have a problem like this you can't solve, try solving an easier version first, like only 2 couples and anything goes.
    $endgroup$
    – DanielV
    13 mins ago
















1












$begingroup$


My book gave the answer as 24. I thought of it like this:



You have four pairs of couples, so you can think of it as M1W2, M2W2, M3W3, M4W4, where
M is a man and W is a woman. M1 has to shake 6 other hands, excluding his wife. You have to do this 4 times for the other men, so you have 4 * 6 handshakes, but in my answer you are double counting. How do I approach this?






combinatorics







share|cite|improve this question









$endgroup$












  • $begingroup$
    In your answer, you both overcounted and undercounted, and incidentally these happened to cancel out and give you the correct answer without having to do anything further. You did $4 times (text{Handshakes done by the men})$, which overcounted the man-man handshakes, but left out the woman-woman handshakes.
    $endgroup$
    – M. Vinay
    38 mins ago










  • $begingroup$
    And that's easily fixed by counting all such handshakes in the same way, not just those done by men, so you get $48$. And now, as you said, you have indeed double-counted. But if you know it's exactly double counting, you can get the answer by halving it!
    $endgroup$
    – M. Vinay
    30 mins ago










  • $begingroup$
    I recommend when you have a problem like this you can't solve, try solving an easier version first, like only 2 couples and anything goes.
    $endgroup$
    – DanielV
    13 mins ago














1












1








1





$begingroup$


My book gave the answer as 24. I thought of it like this:



You have four pairs of couples, so you can think of it as M1W2, M2W2, M3W3, M4W4, where
M is a man and W is a woman. M1 has to shake 6 other hands, excluding his wife. You have to do this 4 times for the other men, so you have 4 * 6 handshakes, but in my answer you are double counting. How do I approach this?






combinatorics







share|cite|improve this question









$endgroup$




My book gave the answer as 24. I thought of it like this:



You have four pairs of couples, so you can think of it as M1W2, M2W2, M3W3, M4W4, where
M is a man and W is a woman. M1 has to shake 6 other hands, excluding his wife. You have to do this 4 times for the other men, so you have 4 * 6 handshakes, but in my answer you are double counting. How do I approach this?






combinatorics




combinatorics






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 45 mins ago









ZakuZaku

592




592












  • $begingroup$
    In your answer, you both overcounted and undercounted, and incidentally these happened to cancel out and give you the correct answer without having to do anything further. You did $4 times (text{Handshakes done by the men})$, which overcounted the man-man handshakes, but left out the woman-woman handshakes.
    $endgroup$
    – M. Vinay
    38 mins ago










  • $begingroup$
    And that's easily fixed by counting all such handshakes in the same way, not just those done by men, so you get $48$. And now, as you said, you have indeed double-counted. But if you know it's exactly double counting, you can get the answer by halving it!
    $endgroup$
    – M. Vinay
    30 mins ago










  • $begingroup$
    I recommend when you have a problem like this you can't solve, try solving an easier version first, like only 2 couples and anything goes.
    $endgroup$
    – DanielV
    13 mins ago


















  • $begingroup$
    In your answer, you both overcounted and undercounted, and incidentally these happened to cancel out and give you the correct answer without having to do anything further. You did $4 times (text{Handshakes done by the men})$, which overcounted the man-man handshakes, but left out the woman-woman handshakes.
    $endgroup$
    – M. Vinay
    38 mins ago










  • $begingroup$
    And that's easily fixed by counting all such handshakes in the same way, not just those done by men, so you get $48$. And now, as you said, you have indeed double-counted. But if you know it's exactly double counting, you can get the answer by halving it!
    $endgroup$
    – M. Vinay
    30 mins ago










  • $begingroup$
    I recommend when you have a problem like this you can't solve, try solving an easier version first, like only 2 couples and anything goes.
    $endgroup$
    – DanielV
    13 mins ago
















$begingroup$
In your answer, you both overcounted and undercounted, and incidentally these happened to cancel out and give you the correct answer without having to do anything further. You did $4 times (text{Handshakes done by the men})$, which overcounted the man-man handshakes, but left out the woman-woman handshakes.
$endgroup$
– M. Vinay
38 mins ago




$begingroup$
In your answer, you both overcounted and undercounted, and incidentally these happened to cancel out and give you the correct answer without having to do anything further. You did $4 times (text{Handshakes done by the men})$, which overcounted the man-man handshakes, but left out the woman-woman handshakes.
$endgroup$
– M. Vinay
38 mins ago












$begingroup$
And that's easily fixed by counting all such handshakes in the same way, not just those done by men, so you get $48$. And now, as you said, you have indeed double-counted. But if you know it's exactly double counting, you can get the answer by halving it!
$endgroup$
– M. Vinay
30 mins ago




$begingroup$
And that's easily fixed by counting all such handshakes in the same way, not just those done by men, so you get $48$. And now, as you said, you have indeed double-counted. But if you know it's exactly double counting, you can get the answer by halving it!
$endgroup$
– M. Vinay
30 mins ago












$begingroup$
I recommend when you have a problem like this you can't solve, try solving an easier version first, like only 2 couples and anything goes.
$endgroup$
– DanielV
13 mins ago




$begingroup$
I recommend when you have a problem like this you can't solve, try solving an easier version first, like only 2 couples and anything goes.
$endgroup$
– DanielV
13 mins ago










3 Answers
3






active

oldest

votes


















2












$begingroup$

Suppose the spouses were allowed to shake each other's hands. That would give you $binom{8}{2} = 28$ handshakes. Since there are four couples, four of these handshakes are illegal. We can remove those to get the $24$ legal handshakes.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    You may proceed as follows using combinations:




    • Number of all possible handshakes among 8 people: $color{blue}{binom{8}{2}}$

    • Number of pairs who do not shake hands: $color{blue}{4}$


    It follows:
    $$mbox{number of hand shakes without pairs} = color{blue}{binom{8}{2}} - color{blue}{4} = frac{8cdot 7}{2} - 4 = 24$$






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      $k$ couples entails $2k$ people. If we imagine each couple going in sequential order, couple 1 will each have to shake $2k-2$ couple's hands for each individual, or $4k-4$ handshakes for couple 1 total. Since there is 1 fewer couple every time a new couple shakes hands, there will be $4k-4i$ handshakes by the $i$-th couple. So the total number of handshakes is given by:



      $$sum_{i=1}^k (4k-4i) = sum_{i=1}^k4k - sum_{i=1}^k4i = 4k^2 - 4frac{k(k+1)}{2} = 4(k^2 - frac{k^2+k}{2}) = 4(k^2 - (frac{k^2}{2} + frac{k}{2})) = 4(frac{k^2}{2}-frac{k}{2}) = 2(k^2-k)$$



      for $k$ couples. Plugging in $k$ = 4 verifies a solution of 24 for this case.






      share|cite|improve this answer










      New contributor




      beefstew2011 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      $endgroup$













      • $begingroup$
        Well, that quite goes against the grain of combinatorics, since you've listed nearly all cases. Try to count without listing. Think lazy!
        $endgroup$
        – M. Vinay
        35 mins ago










      • $begingroup$
        True. I'll delete this.
        $endgroup$
        – beefstew2011
        34 mins ago










      • $begingroup$
        Undeleted with more general answer.
        $endgroup$
        – beefstew2011
        14 mins ago











      Your Answer





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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Suppose the spouses were allowed to shake each other's hands. That would give you $binom{8}{2} = 28$ handshakes. Since there are four couples, four of these handshakes are illegal. We can remove those to get the $24$ legal handshakes.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        Suppose the spouses were allowed to shake each other's hands. That would give you $binom{8}{2} = 28$ handshakes. Since there are four couples, four of these handshakes are illegal. We can remove those to get the $24$ legal handshakes.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          Suppose the spouses were allowed to shake each other's hands. That would give you $binom{8}{2} = 28$ handshakes. Since there are four couples, four of these handshakes are illegal. We can remove those to get the $24$ legal handshakes.






          share|cite|improve this answer









          $endgroup$



          Suppose the spouses were allowed to shake each other's hands. That would give you $binom{8}{2} = 28$ handshakes. Since there are four couples, four of these handshakes are illegal. We can remove those to get the $24$ legal handshakes.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 25 mins ago









          Austin MohrAustin Mohr

          20.5k35098




          20.5k35098























              2












              $begingroup$

              You may proceed as follows using combinations:




              • Number of all possible handshakes among 8 people: $color{blue}{binom{8}{2}}$

              • Number of pairs who do not shake hands: $color{blue}{4}$


              It follows:
              $$mbox{number of hand shakes without pairs} = color{blue}{binom{8}{2}} - color{blue}{4} = frac{8cdot 7}{2} - 4 = 24$$






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                You may proceed as follows using combinations:




                • Number of all possible handshakes among 8 people: $color{blue}{binom{8}{2}}$

                • Number of pairs who do not shake hands: $color{blue}{4}$


                It follows:
                $$mbox{number of hand shakes without pairs} = color{blue}{binom{8}{2}} - color{blue}{4} = frac{8cdot 7}{2} - 4 = 24$$






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  You may proceed as follows using combinations:




                  • Number of all possible handshakes among 8 people: $color{blue}{binom{8}{2}}$

                  • Number of pairs who do not shake hands: $color{blue}{4}$


                  It follows:
                  $$mbox{number of hand shakes without pairs} = color{blue}{binom{8}{2}} - color{blue}{4} = frac{8cdot 7}{2} - 4 = 24$$






                  share|cite|improve this answer









                  $endgroup$



                  You may proceed as follows using combinations:




                  • Number of all possible handshakes among 8 people: $color{blue}{binom{8}{2}}$

                  • Number of pairs who do not shake hands: $color{blue}{4}$


                  It follows:
                  $$mbox{number of hand shakes without pairs} = color{blue}{binom{8}{2}} - color{blue}{4} = frac{8cdot 7}{2} - 4 = 24$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 24 mins ago









                  trancelocationtrancelocation

                  12.6k1826




                  12.6k1826























                      1












                      $begingroup$

                      $k$ couples entails $2k$ people. If we imagine each couple going in sequential order, couple 1 will each have to shake $2k-2$ couple's hands for each individual, or $4k-4$ handshakes for couple 1 total. Since there is 1 fewer couple every time a new couple shakes hands, there will be $4k-4i$ handshakes by the $i$-th couple. So the total number of handshakes is given by:



                      $$sum_{i=1}^k (4k-4i) = sum_{i=1}^k4k - sum_{i=1}^k4i = 4k^2 - 4frac{k(k+1)}{2} = 4(k^2 - frac{k^2+k}{2}) = 4(k^2 - (frac{k^2}{2} + frac{k}{2})) = 4(frac{k^2}{2}-frac{k}{2}) = 2(k^2-k)$$



                      for $k$ couples. Plugging in $k$ = 4 verifies a solution of 24 for this case.






                      share|cite|improve this answer










                      New contributor




                      beefstew2011 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                      Check out our Code of Conduct.






                      $endgroup$













                      • $begingroup$
                        Well, that quite goes against the grain of combinatorics, since you've listed nearly all cases. Try to count without listing. Think lazy!
                        $endgroup$
                        – M. Vinay
                        35 mins ago










                      • $begingroup$
                        True. I'll delete this.
                        $endgroup$
                        – beefstew2011
                        34 mins ago










                      • $begingroup$
                        Undeleted with more general answer.
                        $endgroup$
                        – beefstew2011
                        14 mins ago
















                      1












                      $begingroup$

                      $k$ couples entails $2k$ people. If we imagine each couple going in sequential order, couple 1 will each have to shake $2k-2$ couple's hands for each individual, or $4k-4$ handshakes for couple 1 total. Since there is 1 fewer couple every time a new couple shakes hands, there will be $4k-4i$ handshakes by the $i$-th couple. So the total number of handshakes is given by:



                      $$sum_{i=1}^k (4k-4i) = sum_{i=1}^k4k - sum_{i=1}^k4i = 4k^2 - 4frac{k(k+1)}{2} = 4(k^2 - frac{k^2+k}{2}) = 4(k^2 - (frac{k^2}{2} + frac{k}{2})) = 4(frac{k^2}{2}-frac{k}{2}) = 2(k^2-k)$$



                      for $k$ couples. Plugging in $k$ = 4 verifies a solution of 24 for this case.






                      share|cite|improve this answer










                      New contributor




                      beefstew2011 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                      Check out our Code of Conduct.






                      $endgroup$













                      • $begingroup$
                        Well, that quite goes against the grain of combinatorics, since you've listed nearly all cases. Try to count without listing. Think lazy!
                        $endgroup$
                        – M. Vinay
                        35 mins ago










                      • $begingroup$
                        True. I'll delete this.
                        $endgroup$
                        – beefstew2011
                        34 mins ago










                      • $begingroup$
                        Undeleted with more general answer.
                        $endgroup$
                        – beefstew2011
                        14 mins ago














                      1












                      1








                      1





                      $begingroup$

                      $k$ couples entails $2k$ people. If we imagine each couple going in sequential order, couple 1 will each have to shake $2k-2$ couple's hands for each individual, or $4k-4$ handshakes for couple 1 total. Since there is 1 fewer couple every time a new couple shakes hands, there will be $4k-4i$ handshakes by the $i$-th couple. So the total number of handshakes is given by:



                      $$sum_{i=1}^k (4k-4i) = sum_{i=1}^k4k - sum_{i=1}^k4i = 4k^2 - 4frac{k(k+1)}{2} = 4(k^2 - frac{k^2+k}{2}) = 4(k^2 - (frac{k^2}{2} + frac{k}{2})) = 4(frac{k^2}{2}-frac{k}{2}) = 2(k^2-k)$$



                      for $k$ couples. Plugging in $k$ = 4 verifies a solution of 24 for this case.






                      share|cite|improve this answer










                      New contributor




                      beefstew2011 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                      Check out our Code of Conduct.






                      $endgroup$



                      $k$ couples entails $2k$ people. If we imagine each couple going in sequential order, couple 1 will each have to shake $2k-2$ couple's hands for each individual, or $4k-4$ handshakes for couple 1 total. Since there is 1 fewer couple every time a new couple shakes hands, there will be $4k-4i$ handshakes by the $i$-th couple. So the total number of handshakes is given by:



                      $$sum_{i=1}^k (4k-4i) = sum_{i=1}^k4k - sum_{i=1}^k4i = 4k^2 - 4frac{k(k+1)}{2} = 4(k^2 - frac{k^2+k}{2}) = 4(k^2 - (frac{k^2}{2} + frac{k}{2})) = 4(frac{k^2}{2}-frac{k}{2}) = 2(k^2-k)$$



                      for $k$ couples. Plugging in $k$ = 4 verifies a solution of 24 for this case.







                      share|cite|improve this answer










                      New contributor




                      beefstew2011 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                      Check out our Code of Conduct.









                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited 14 mins ago





















                      New contributor




                      beefstew2011 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                      Check out our Code of Conduct.









                      answered 37 mins ago









                      beefstew2011beefstew2011

                      687




                      687




                      New contributor




                      beefstew2011 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                      Check out our Code of Conduct.





                      New contributor





                      beefstew2011 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                      Check out our Code of Conduct.






                      beefstew2011 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                      Check out our Code of Conduct.












                      • $begingroup$
                        Well, that quite goes against the grain of combinatorics, since you've listed nearly all cases. Try to count without listing. Think lazy!
                        $endgroup$
                        – M. Vinay
                        35 mins ago










                      • $begingroup$
                        True. I'll delete this.
                        $endgroup$
                        – beefstew2011
                        34 mins ago










                      • $begingroup$
                        Undeleted with more general answer.
                        $endgroup$
                        – beefstew2011
                        14 mins ago


















                      • $begingroup$
                        Well, that quite goes against the grain of combinatorics, since you've listed nearly all cases. Try to count without listing. Think lazy!
                        $endgroup$
                        – M. Vinay
                        35 mins ago










                      • $begingroup$
                        True. I'll delete this.
                        $endgroup$
                        – beefstew2011
                        34 mins ago










                      • $begingroup$
                        Undeleted with more general answer.
                        $endgroup$
                        – beefstew2011
                        14 mins ago
















                      $begingroup$
                      Well, that quite goes against the grain of combinatorics, since you've listed nearly all cases. Try to count without listing. Think lazy!
                      $endgroup$
                      – M. Vinay
                      35 mins ago




                      $begingroup$
                      Well, that quite goes against the grain of combinatorics, since you've listed nearly all cases. Try to count without listing. Think lazy!
                      $endgroup$
                      – M. Vinay
                      35 mins ago












                      $begingroup$
                      True. I'll delete this.
                      $endgroup$
                      – beefstew2011
                      34 mins ago




                      $begingroup$
                      True. I'll delete this.
                      $endgroup$
                      – beefstew2011
                      34 mins ago












                      $begingroup$
                      Undeleted with more general answer.
                      $endgroup$
                      – beefstew2011
                      14 mins ago




                      $begingroup$
                      Undeleted with more general answer.
                      $endgroup$
                      – beefstew2011
                      14 mins ago


















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