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
    49 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
    41 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
    24 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
    6 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
    49 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
    41 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
    24 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
    6 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 57 mins 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
    49 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
    41 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
    24 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
    6 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
    49 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
    41 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
    24 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
    6 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
49 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
49 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
41 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
41 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
24 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
24 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
6 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
6 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
        46 mins ago










      • $begingroup$
        Undeleted with more general answer.
        $endgroup$
        – beefstew2011
        25 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
        2 mins ago











      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3151152%2ffour-married-couples-attend-a-party-each-person-shakes-hands-with-every-other-p%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      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 36 mins 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 35 mins 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
                        46 mins ago










                      • $begingroup$
                        Undeleted with more general answer.
                        $endgroup$
                        – beefstew2011
                        25 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
                        2 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
                        46 mins ago










                      • $begingroup$
                        Undeleted with more general answer.
                        $endgroup$
                        – beefstew2011
                        25 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
                        2 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 26 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 48 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.












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










                      • $begingroup$
                        Undeleted with more general answer.
                        $endgroup$
                        – beefstew2011
                        25 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
                        2 mins ago


















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










                      • $begingroup$
                        Undeleted with more general answer.
                        $endgroup$
                        – beefstew2011
                        25 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
                        2 mins ago
















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




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












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




                      $begingroup$
                      Undeleted with more general answer.
                      $endgroup$
                      – beefstew2011
                      25 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
                      2 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
                      2 mins ago


















                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3151152%2ffour-married-couples-attend-a-party-each-person-shakes-hands-with-every-other-p%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      -

                      Popular posts from this blog

                      Knooppunt Holsloot

                      Image
                      Knooppunt Holsloot Kenmerken Wegen A37 × N34 Type knooppunt klaverblad Portaal     Verkeer & Vervoer Knooppunt Holsloot is een verkeersknooppunt op de kruising van de autosnelweg A37 en de N34 in Drenthe, tussen Emmen en Coevorden. Oorspronkelijk was deze aansluiting een afritconstructie van het type Haarlemmermeeraansluiting, zoals bij Knooppunt Hooipolder. Sinds 2002 is het knooppunt omgebouwd tot een volledig klaverblad. Als noviteit werd hier voor het eerst gebruikgemaakt van dimbare verlichting in plaats van de normale vaste tijden waarop de verlichting aan en uit gaat, wat een grote kostenbesparing opleverde. [1] Dat Holsloot als volwaardig knooppunt is uitgevoerd, waarbij alle verkeersstromen en afbuigingen ononderbroken kunnen doorgaan, is uitzonderlijk. Het betreft hier namelijk een aansluiting tussen een autosnelweg en niet-snelweg. Een vergelijkbare situatie treft men nabij Hoensbroek bij het knooppunt Ten Esschen.     Aa
                      Read more

                      Altaar (religie)

                      Image
                      Deel van de serie over kerkelijk gerei waaronder paramenten en liturgisch vaatwerk gebruikt in de liturgie Liturgisch vaatwerk Vasa sacra Miskelk · Pateen Kelklepeltje Ciborie · Monstrans Pyxis · Custodiale Vasa non sacra Ampullen · Wijwatervat Olievaatje Wierookvat · Ablutievat Paramenten Amict · Albe · Baarkleed · Cingel Tuniek · Dalmatiek · Fanon Kazuifel · Manipel · Stola Gremiale · Benedictievelum Mijter Koorkledij Rochet · Superplie Koorkap · Cappa magna Kovel Kelkgerei Bursa · Kelkvelum Ciborievelum Kerklinnen Corporale · Kelkdoekje Lavabodoekje · Palla Altaardwaal Kerkinterieur Altaar · Ambo Biechtstoel · Communiebank Doksaal · Doopvont Faldistorium Godslamp · Hoogaltaar Heilig Kruisaltaar Katheder · Preekstoel Sedilia Tabernakel · Volksaltaar Liturgische boeken Altaarmissaal Benedictionale · Brevier Evangeliarium · Evangelistarium Graduale · Kyriale Lectionarium Psalter · Rituaal Sacramentarium · Volksmissaal Overige Flambou
                      Read more

                      Gregoriusmis

                      Image
                      Een afbeelding van de Gregoriusmis uit een getijdenboek gemaakt in de zuidelijke Nederlanden tussen 1500 en 1525 Gregoriusmis, Meester van het Akense Altaar, ca. 1500-1525. Collectie Museum Catharijneconvent De Gregoriusmis of de Mis van Sint Gregorius is een motief in katholieke kunst, dat het eerst verscheen in de late Middeleeuwen en dat nog steeds werd gebruikt tijdens de contrareformatie. Paus Gregorius I (ca. 540-604) is afgebeeld, de H. Mis opdragend, terwijl Christus als de Man van Smarten verschijnt op het altaar dat voor Sint Gregorius staat. Volgens de legende had St. Gregorius gebeden om een teken voor een betwijfelaar van de transsubstantiatieleer, waarop de lijdende Christus tijdens het opdragen van de Mis verscheen om hem te overtuigen en waarbij zijn bloed in de altaarkelk liep. Vaak zijn ook de passiewerktuigen afgebeeld. Zie de categorie Mass of Saint Gregory van Wikimedia Commons voor mediabestanden over dit onderwerp. This page is onl
                      Read more