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












2












$begingroup$


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
    1 hour 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
    1 hour 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
    1 hour ago










  • $begingroup$
    Only person #1 has to shake hands 6 times, person #2 has already shaken hands with Person #1, so he only has to shake hands with 5 people. So the answer becomes 6+5+4+3+2+1, or 21. So Yes, I believe 21 is correct, to prevent double counting.
    $endgroup$
    – Issel
    43 mins ago










  • $begingroup$
    @Issel No, Person #2 being the spouse of Person #1, also has to shake hands with $6$ people, and so on, so it's $6 + 6 + 4 + 4 + 2 + 2 + 0 + 0 = 24$.
    $endgroup$
    – M. Vinay
    33 mins ago
















2












$begingroup$


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
    1 hour 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
    1 hour 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
    1 hour ago










  • $begingroup$
    Only person #1 has to shake hands 6 times, person #2 has already shaken hands with Person #1, so he only has to shake hands with 5 people. So the answer becomes 6+5+4+3+2+1, or 21. So Yes, I believe 21 is correct, to prevent double counting.
    $endgroup$
    – Issel
    43 mins ago










  • $begingroup$
    @Issel No, Person #2 being the spouse of Person #1, also has to shake hands with $6$ people, and so on, so it's $6 + 6 + 4 + 4 + 2 + 2 + 0 + 0 = 24$.
    $endgroup$
    – M. Vinay
    33 mins ago














2












2








2





$begingroup$


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 1 hour ago









ZakuZaku

642




642












  • $begingroup$
    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
    1 hour 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
    1 hour 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
    1 hour ago










  • $begingroup$
    Only person #1 has to shake hands 6 times, person #2 has already shaken hands with Person #1, so he only has to shake hands with 5 people. So the answer becomes 6+5+4+3+2+1, or 21. So Yes, I believe 21 is correct, to prevent double counting.
    $endgroup$
    – Issel
    43 mins ago










  • $begingroup$
    @Issel No, Person #2 being the spouse of Person #1, also has to shake hands with $6$ people, and so on, so it's $6 + 6 + 4 + 4 + 2 + 2 + 0 + 0 = 24$.
    $endgroup$
    – M. Vinay
    33 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
    1 hour 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
    1 hour 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
    1 hour ago










  • $begingroup$
    Only person #1 has to shake hands 6 times, person #2 has already shaken hands with Person #1, so he only has to shake hands with 5 people. So the answer becomes 6+5+4+3+2+1, or 21. So Yes, I believe 21 is correct, to prevent double counting.
    $endgroup$
    – Issel
    43 mins ago










  • $begingroup$
    @Issel No, Person #2 being the spouse of Person #1, also has to shake hands with $6$ people, and so on, so it's $6 + 6 + 4 + 4 + 2 + 2 + 0 + 0 = 24$.
    $endgroup$
    – M. Vinay
    33 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
1 hour 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
1 hour 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
1 hour 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
1 hour 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
1 hour 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
1 hour ago












$begingroup$
Only person #1 has to shake hands 6 times, person #2 has already shaken hands with Person #1, so he only has to shake hands with 5 people. So the answer becomes 6+5+4+3+2+1, or 21. So Yes, I believe 21 is correct, to prevent double counting.
$endgroup$
– Issel
43 mins ago




$begingroup$
Only person #1 has to shake hands 6 times, person #2 has already shaken hands with Person #1, so he only has to shake hands with 5 people. So the answer becomes 6+5+4+3+2+1, or 21. So Yes, I believe 21 is correct, to prevent double counting.
$endgroup$
– Issel
43 mins ago












$begingroup$
@Issel No, Person #2 being the spouse of Person #1, also has to shake hands with $6$ people, and so on, so it's $6 + 6 + 4 + 4 + 2 + 2 + 0 + 0 = 24$.
$endgroup$
– M. Vinay
33 mins ago




$begingroup$
@Issel No, Person #2 being the spouse of Person #1, also has to shake hands with $6$ people, and so on, so it's $6 + 6 + 4 + 4 + 2 + 2 + 0 + 0 = 24$.
$endgroup$
– M. Vinay
33 mins ago










3 Answers
3






active

oldest

votes


















4












$begingroup$

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









$endgroup$





















    4












    $begingroup$

    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.






      $endgroup$













      • $begingroup$
        True. I'll delete this.
        $endgroup$
        – beefstew2011
        1 hour ago










      • $begingroup$
        Undeleted with more general answer.
        $endgroup$
        – beefstew2011
        1 hour ago










      • $begingroup$
        Well… Each of the $2k$ people shakes hands with $2k - 1 - 1 = 2k - 2$ others (everyone except the spouse). So that's $2k(2k- 2) = 4k(k - 1)$, but since every handshake must've been counted twice, divide that by $2$ to get $2k(k - 1)$ handshakes in total.
        $endgroup$
        – M. Vinay
        39 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









      4












      $begingroup$

      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









      $endgroup$


















        4












        $begingroup$

        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









        $endgroup$
















          4












          4








          4





          $begingroup$

          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 1 hour ago









          Austin MohrAustin Mohr

          20.5k35098




          20.5k35098























              4












              $begingroup$

              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









              $endgroup$


















                4












                $begingroup$

                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









                $endgroup$
















                  4












                  4








                  4





                  $begingroup$

                  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 1 hour ago









                  trancelocationtrancelocation

                  12.7k1826




                  12.7k1826























                      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.






                      $endgroup$













                      • $begingroup$
                        True. I'll delete this.
                        $endgroup$
                        – beefstew2011
                        1 hour ago










                      • $begingroup$
                        Undeleted with more general answer.
                        $endgroup$
                        – beefstew2011
                        1 hour ago










                      • $begingroup$
                        Well… Each of the $2k$ people shakes hands with $2k - 1 - 1 = 2k - 2$ others (everyone except the spouse). So that's $2k(2k- 2) = 4k(k - 1)$, but since every handshake must've been counted twice, divide that by $2$ to get $2k(k - 1)$ handshakes in total.
                        $endgroup$
                        – M. Vinay
                        39 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.






                      $endgroup$













                      • $begingroup$
                        True. I'll delete this.
                        $endgroup$
                        – beefstew2011
                        1 hour ago










                      • $begingroup$
                        Undeleted with more general answer.
                        $endgroup$
                        – beefstew2011
                        1 hour ago










                      • $begingroup$
                        Well… Each of the $2k$ people shakes hands with $2k - 1 - 1 = 2k - 2$ others (everyone except the spouse). So that's $2k(2k- 2) = 4k(k - 1)$, but since every handshake must've been counted twice, divide that by $2$ to get $2k(k - 1)$ handshakes in total.
                        $endgroup$
                        – M. Vinay
                        39 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 1 hour 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 1 hour 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.












                      • $begingroup$
                        True. I'll delete this.
                        $endgroup$
                        – beefstew2011
                        1 hour ago










                      • $begingroup$
                        Undeleted with more general answer.
                        $endgroup$
                        – beefstew2011
                        1 hour ago










                      • $begingroup$
                        Well… Each of the $2k$ people shakes hands with $2k - 1 - 1 = 2k - 2$ others (everyone except the spouse). So that's $2k(2k- 2) = 4k(k - 1)$, but since every handshake must've been counted twice, divide that by $2$ to get $2k(k - 1)$ handshakes in total.
                        $endgroup$
                        – M. Vinay
                        39 mins ago


















                      • $begingroup$
                        True. I'll delete this.
                        $endgroup$
                        – beefstew2011
                        1 hour ago










                      • $begingroup$
                        Undeleted with more general answer.
                        $endgroup$
                        – beefstew2011
                        1 hour ago










                      • $begingroup$
                        Well… Each of the $2k$ people shakes hands with $2k - 1 - 1 = 2k - 2$ others (everyone except the spouse). So that's $2k(2k- 2) = 4k(k - 1)$, but since every handshake must've been counted twice, divide that by $2$ to get $2k(k - 1)$ handshakes in total.
                        $endgroup$
                        – M. Vinay
                        39 mins ago
















                      $begingroup$
                      True. I'll delete this.
                      $endgroup$
                      – beefstew2011
                      1 hour ago




                      $begingroup$
                      True. I'll delete this.
                      $endgroup$
                      – beefstew2011
                      1 hour ago












                      $begingroup$
                      Undeleted with more general answer.
                      $endgroup$
                      – beefstew2011
                      1 hour ago




                      $begingroup$
                      Undeleted with more general answer.
                      $endgroup$
                      – beefstew2011
                      1 hour ago












                      $begingroup$
                      Well… Each of the $2k$ people shakes hands with $2k - 1 - 1 = 2k - 2$ others (everyone except the spouse). So that's $2k(2k- 2) = 4k(k - 1)$, but since every handshake must've been counted twice, divide that by $2$ to get $2k(k - 1)$ handshakes in total.
                      $endgroup$
                      – M. Vinay
                      39 mins ago




                      $begingroup$
                      Well… Each of the $2k$ people shakes hands with $2k - 1 - 1 = 2k - 2$ others (everyone except the spouse). So that's $2k(2k- 2) = 4k(k - 1)$, but since every handshake must've been counted twice, divide that by $2$ to get $2k(k - 1)$ handshakes in total.
                      $endgroup$
                      – M. Vinay
                      39 mins ago


















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