A dance class consists of 22 students, 10 women and 12 men. If 5 men and 5 women are to be chosen and then...












1












$begingroup$


This is a question from Sheldon Ross.




A dance class consists of 22 students, 10 women and 12 men. If
5 men and 5 women are to be chosen and then paired off, how many results are possible?




So my reasoning is this :




  1. I choose 5 women from a pool of 10 in 10C2 ways.

  2. I choose 5 men from a pool of 12 in 12C2 ways.


So total number of ways of choosing in 10C2 x 12C2. Now I need to arrange them in 5 pairs. This is where I have a different solution. The solution says that there are 5! ways to arrange them in pairs.



But I cant seem to understand why? My reasoning is that for first pair position I need to choose 1 man from 5 and 1 woman from 5. So for the first position I have 5 x 5 choices (5 for man and 5 for woman). Similarly for the second position I have 4 x 4 choices and so on. Hence the total ways are 5! x 5!



So I calculate the total ways as 10C2 * 12C2 * 5! * 5!. Can anyone point the flaw in my reasoning for arranging the chosen men and women in pairs.



Thanks in advance






combinatorics permutations combinations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Please make your titles informative. We get dozens of "permutations combinations" questions everyday.
    $endgroup$
    – Em.
    Jul 10 '16 at 8:28












  • $begingroup$
    Well, to pair up, the first B/G can choose 5G/B, then 4 and so on, thus it is 5!, not $5!^2$
    $endgroup$
    – Ariana
    Jul 10 '16 at 8:53






  • 3




    $begingroup$
    Did you mean $binom{10}{5}binom{12}{5}5!$?
    $endgroup$
    – N. F. Taussig
    Jul 10 '16 at 9:08










  • $begingroup$
    @N.F.Taussig, Yes that's what I meant. That's the answer in book. My reasoning has one more 5!.
    $endgroup$
    – amrx
    Jul 10 '16 at 9:17






  • 1




    $begingroup$
    As Mike indicated in his answer, once you have selected five men and five women, you can line up the men in some order, say alphabetically. The first man in the list can be matched with any of the five women, the second man can be matched with one of the four remaining women, and so forth. By multiplying $binom{10}{5}binom{12}{5}5!$ by $5!$, you are saying that the order in which the couples are selected matters.
    $endgroup$
    – N. F. Taussig
    Jul 10 '16 at 9:24
















1












$begingroup$


This is a question from Sheldon Ross.




A dance class consists of 22 students, 10 women and 12 men. If
5 men and 5 women are to be chosen and then paired off, how many results are possible?




So my reasoning is this :




  1. I choose 5 women from a pool of 10 in 10C2 ways.

  2. I choose 5 men from a pool of 12 in 12C2 ways.


So total number of ways of choosing in 10C2 x 12C2. Now I need to arrange them in 5 pairs. This is where I have a different solution. The solution says that there are 5! ways to arrange them in pairs.



But I cant seem to understand why? My reasoning is that for first pair position I need to choose 1 man from 5 and 1 woman from 5. So for the first position I have 5 x 5 choices (5 for man and 5 for woman). Similarly for the second position I have 4 x 4 choices and so on. Hence the total ways are 5! x 5!



So I calculate the total ways as 10C2 * 12C2 * 5! * 5!. Can anyone point the flaw in my reasoning for arranging the chosen men and women in pairs.



Thanks in advance






combinatorics permutations combinations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Please make your titles informative. We get dozens of "permutations combinations" questions everyday.
    $endgroup$
    – Em.
    Jul 10 '16 at 8:28












  • $begingroup$
    Well, to pair up, the first B/G can choose 5G/B, then 4 and so on, thus it is 5!, not $5!^2$
    $endgroup$
    – Ariana
    Jul 10 '16 at 8:53






  • 3




    $begingroup$
    Did you mean $binom{10}{5}binom{12}{5}5!$?
    $endgroup$
    – N. F. Taussig
    Jul 10 '16 at 9:08










  • $begingroup$
    @N.F.Taussig, Yes that's what I meant. That's the answer in book. My reasoning has one more 5!.
    $endgroup$
    – amrx
    Jul 10 '16 at 9:17






  • 1




    $begingroup$
    As Mike indicated in his answer, once you have selected five men and five women, you can line up the men in some order, say alphabetically. The first man in the list can be matched with any of the five women, the second man can be matched with one of the four remaining women, and so forth. By multiplying $binom{10}{5}binom{12}{5}5!$ by $5!$, you are saying that the order in which the couples are selected matters.
    $endgroup$
    – N. F. Taussig
    Jul 10 '16 at 9:24














1












1








1





$begingroup$


This is a question from Sheldon Ross.




A dance class consists of 22 students, 10 women and 12 men. If
5 men and 5 women are to be chosen and then paired off, how many results are possible?




So my reasoning is this :




  1. I choose 5 women from a pool of 10 in 10C2 ways.

  2. I choose 5 men from a pool of 12 in 12C2 ways.


So total number of ways of choosing in 10C2 x 12C2. Now I need to arrange them in 5 pairs. This is where I have a different solution. The solution says that there are 5! ways to arrange them in pairs.



But I cant seem to understand why? My reasoning is that for first pair position I need to choose 1 man from 5 and 1 woman from 5. So for the first position I have 5 x 5 choices (5 for man and 5 for woman). Similarly for the second position I have 4 x 4 choices and so on. Hence the total ways are 5! x 5!



So I calculate the total ways as 10C2 * 12C2 * 5! * 5!. Can anyone point the flaw in my reasoning for arranging the chosen men and women in pairs.



Thanks in advance






combinatorics permutations combinations







share|cite|improve this question











$endgroup$




This is a question from Sheldon Ross.




A dance class consists of 22 students, 10 women and 12 men. If
5 men and 5 women are to be chosen and then paired off, how many results are possible?




So my reasoning is this :




  1. I choose 5 women from a pool of 10 in 10C2 ways.

  2. I choose 5 men from a pool of 12 in 12C2 ways.


So total number of ways of choosing in 10C2 x 12C2. Now I need to arrange them in 5 pairs. This is where I have a different solution. The solution says that there are 5! ways to arrange them in pairs.



But I cant seem to understand why? My reasoning is that for first pair position I need to choose 1 man from 5 and 1 woman from 5. So for the first position I have 5 x 5 choices (5 for man and 5 for woman). Similarly for the second position I have 4 x 4 choices and so on. Hence the total ways are 5! x 5!



So I calculate the total ways as 10C2 * 12C2 * 5! * 5!. Can anyone point the flaw in my reasoning for arranging the chosen men and women in pairs.



Thanks in advance






combinatorics permutations combinations




combinatorics permutations combinations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jul 10 '16 at 8:28









Em.

15k72037




15k72037










asked Jul 10 '16 at 8:26









amrxamrx

775




775












  • $begingroup$
    Please make your titles informative. We get dozens of "permutations combinations" questions everyday.
    $endgroup$
    – Em.
    Jul 10 '16 at 8:28












  • $begingroup$
    Well, to pair up, the first B/G can choose 5G/B, then 4 and so on, thus it is 5!, not $5!^2$
    $endgroup$
    – Ariana
    Jul 10 '16 at 8:53






  • 3




    $begingroup$
    Did you mean $binom{10}{5}binom{12}{5}5!$?
    $endgroup$
    – N. F. Taussig
    Jul 10 '16 at 9:08










  • $begingroup$
    @N.F.Taussig, Yes that's what I meant. That's the answer in book. My reasoning has one more 5!.
    $endgroup$
    – amrx
    Jul 10 '16 at 9:17






  • 1




    $begingroup$
    As Mike indicated in his answer, once you have selected five men and five women, you can line up the men in some order, say alphabetically. The first man in the list can be matched with any of the five women, the second man can be matched with one of the four remaining women, and so forth. By multiplying $binom{10}{5}binom{12}{5}5!$ by $5!$, you are saying that the order in which the couples are selected matters.
    $endgroup$
    – N. F. Taussig
    Jul 10 '16 at 9:24


















  • $begingroup$
    Please make your titles informative. We get dozens of "permutations combinations" questions everyday.
    $endgroup$
    – Em.
    Jul 10 '16 at 8:28












  • $begingroup$
    Well, to pair up, the first B/G can choose 5G/B, then 4 and so on, thus it is 5!, not $5!^2$
    $endgroup$
    – Ariana
    Jul 10 '16 at 8:53






  • 3




    $begingroup$
    Did you mean $binom{10}{5}binom{12}{5}5!$?
    $endgroup$
    – N. F. Taussig
    Jul 10 '16 at 9:08










  • $begingroup$
    @N.F.Taussig, Yes that's what I meant. That's the answer in book. My reasoning has one more 5!.
    $endgroup$
    – amrx
    Jul 10 '16 at 9:17






  • 1




    $begingroup$
    As Mike indicated in his answer, once you have selected five men and five women, you can line up the men in some order, say alphabetically. The first man in the list can be matched with any of the five women, the second man can be matched with one of the four remaining women, and so forth. By multiplying $binom{10}{5}binom{12}{5}5!$ by $5!$, you are saying that the order in which the couples are selected matters.
    $endgroup$
    – N. F. Taussig
    Jul 10 '16 at 9:24
















$begingroup$
Please make your titles informative. We get dozens of "permutations combinations" questions everyday.
$endgroup$
– Em.
Jul 10 '16 at 8:28






$begingroup$
Please make your titles informative. We get dozens of "permutations combinations" questions everyday.
$endgroup$
– Em.
Jul 10 '16 at 8:28














$begingroup$
Well, to pair up, the first B/G can choose 5G/B, then 4 and so on, thus it is 5!, not $5!^2$
$endgroup$
– Ariana
Jul 10 '16 at 8:53




$begingroup$
Well, to pair up, the first B/G can choose 5G/B, then 4 and so on, thus it is 5!, not $5!^2$
$endgroup$
– Ariana
Jul 10 '16 at 8:53




3




3




$begingroup$
Did you mean $binom{10}{5}binom{12}{5}5!$?
$endgroup$
– N. F. Taussig
Jul 10 '16 at 9:08




$begingroup$
Did you mean $binom{10}{5}binom{12}{5}5!$?
$endgroup$
– N. F. Taussig
Jul 10 '16 at 9:08












$begingroup$
@N.F.Taussig, Yes that's what I meant. That's the answer in book. My reasoning has one more 5!.
$endgroup$
– amrx
Jul 10 '16 at 9:17




$begingroup$
@N.F.Taussig, Yes that's what I meant. That's the answer in book. My reasoning has one more 5!.
$endgroup$
– amrx
Jul 10 '16 at 9:17




1




1




$begingroup$
As Mike indicated in his answer, once you have selected five men and five women, you can line up the men in some order, say alphabetically. The first man in the list can be matched with any of the five women, the second man can be matched with one of the four remaining women, and so forth. By multiplying $binom{10}{5}binom{12}{5}5!$ by $5!$, you are saying that the order in which the couples are selected matters.
$endgroup$
– N. F. Taussig
Jul 10 '16 at 9:24




$begingroup$
As Mike indicated in his answer, once you have selected five men and five women, you can line up the men in some order, say alphabetically. The first man in the list can be matched with any of the five women, the second man can be matched with one of the four remaining women, and so forth. By multiplying $binom{10}{5}binom{12}{5}5!$ by $5!$, you are saying that the order in which the couples are selected matters.
$endgroup$
– N. F. Taussig
Jul 10 '16 at 9:24










2 Answers
2






active

oldest

votes


















2












$begingroup$

I believe that the problem is that you're counting multiple times. If the couples were ordered (first couple, second couple, etc.) then it would be correct. In this case, though, you can simply go down the list of men and ask yourself "How many different women can I pair with this man?"






share|cite|improve this answer









$endgroup$









  • 3




    $begingroup$
    As a footnote, interestingly enough, this problem doesn't seem to explicitly forbid male/male or female/female couples, which would change the answer entirely.
    $endgroup$
    – Mike
    Jul 10 '16 at 8:35



















0












$begingroup$

After selecting 5 from men and 5 from women, we need to pair them.



Lets assume like finding a pair for each man,



for the 1st guy -- can choose 1 from 5 women



for the 2nd guy -- can choose 1 from 4 women



.



.



for the 5th guy -- can choose 1 from 1 woman



so 5!






share|cite|improve this answer









$endgroup$














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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    I believe that the problem is that you're counting multiple times. If the couples were ordered (first couple, second couple, etc.) then it would be correct. In this case, though, you can simply go down the list of men and ask yourself "How many different women can I pair with this man?"






    share|cite|improve this answer









    $endgroup$









    • 3




      $begingroup$
      As a footnote, interestingly enough, this problem doesn't seem to explicitly forbid male/male or female/female couples, which would change the answer entirely.
      $endgroup$
      – Mike
      Jul 10 '16 at 8:35
















    2












    $begingroup$

    I believe that the problem is that you're counting multiple times. If the couples were ordered (first couple, second couple, etc.) then it would be correct. In this case, though, you can simply go down the list of men and ask yourself "How many different women can I pair with this man?"






    share|cite|improve this answer









    $endgroup$









    • 3




      $begingroup$
      As a footnote, interestingly enough, this problem doesn't seem to explicitly forbid male/male or female/female couples, which would change the answer entirely.
      $endgroup$
      – Mike
      Jul 10 '16 at 8:35














    2












    2








    2





    $begingroup$

    I believe that the problem is that you're counting multiple times. If the couples were ordered (first couple, second couple, etc.) then it would be correct. In this case, though, you can simply go down the list of men and ask yourself "How many different women can I pair with this man?"






    share|cite|improve this answer









    $endgroup$



    I believe that the problem is that you're counting multiple times. If the couples were ordered (first couple, second couple, etc.) then it would be correct. In this case, though, you can simply go down the list of men and ask yourself "How many different women can I pair with this man?"







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jul 10 '16 at 8:33









    MikeMike

    11.6k31743




    11.6k31743








    • 3




      $begingroup$
      As a footnote, interestingly enough, this problem doesn't seem to explicitly forbid male/male or female/female couples, which would change the answer entirely.
      $endgroup$
      – Mike
      Jul 10 '16 at 8:35














    • 3




      $begingroup$
      As a footnote, interestingly enough, this problem doesn't seem to explicitly forbid male/male or female/female couples, which would change the answer entirely.
      $endgroup$
      – Mike
      Jul 10 '16 at 8:35








    3




    3




    $begingroup$
    As a footnote, interestingly enough, this problem doesn't seem to explicitly forbid male/male or female/female couples, which would change the answer entirely.
    $endgroup$
    – Mike
    Jul 10 '16 at 8:35




    $begingroup$
    As a footnote, interestingly enough, this problem doesn't seem to explicitly forbid male/male or female/female couples, which would change the answer entirely.
    $endgroup$
    – Mike
    Jul 10 '16 at 8:35











    0












    $begingroup$

    After selecting 5 from men and 5 from women, we need to pair them.



    Lets assume like finding a pair for each man,



    for the 1st guy -- can choose 1 from 5 women



    for the 2nd guy -- can choose 1 from 4 women



    .



    .



    for the 5th guy -- can choose 1 from 1 woman



    so 5!






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      After selecting 5 from men and 5 from women, we need to pair them.



      Lets assume like finding a pair for each man,



      for the 1st guy -- can choose 1 from 5 women



      for the 2nd guy -- can choose 1 from 4 women



      .



      .



      for the 5th guy -- can choose 1 from 1 woman



      so 5!






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        After selecting 5 from men and 5 from women, we need to pair them.



        Lets assume like finding a pair for each man,



        for the 1st guy -- can choose 1 from 5 women



        for the 2nd guy -- can choose 1 from 4 women



        .



        .



        for the 5th guy -- can choose 1 from 1 woman



        so 5!






        share|cite|improve this answer









        $endgroup$



        After selecting 5 from men and 5 from women, we need to pair them.



        Lets assume like finding a pair for each man,



        for the 1st guy -- can choose 1 from 5 women



        for the 2nd guy -- can choose 1 from 4 women



        .



        .



        for the 5th guy -- can choose 1 from 1 woman



        so 5!







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 28 '18 at 13:17









        loyolaloyola

        1011




        1011






























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