Number of ways to pick an equal number of elements from two sets?











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There are two groups such that one group contains $m$ elements and another group contains $n$ elements. We have to find number of ways to pick same number of elements from both the groups.




My approach is
$1+dbinom m1.dbinom n1+dbinom m2 . dbinom n2 +.... $upto $dbinom mm$ or $dbinom nn$ whatever is smaller.

Is there any faster way to do the same?



Example:
$3$ and $2$
$1+dbinom 31.dbinom 21+dbinom 32 . dbinom 22=10$ ways






combinatorics permutations







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    up vote
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    down vote

    favorite













    There are two groups such that one group contains $m$ elements and another group contains $n$ elements. We have to find number of ways to pick same number of elements from both the groups.




    My approach is
    $1+dbinom m1.dbinom n1+dbinom m2 . dbinom n2 +.... $upto $dbinom mm$ or $dbinom nn$ whatever is smaller.

    Is there any faster way to do the same?



    Example:
    $3$ and $2$
    $1+dbinom 31.dbinom 21+dbinom 32 . dbinom 22=10$ ways






    combinatorics permutations







    share|cite|improve this question









    New contributor




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






















      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite












      There are two groups such that one group contains $m$ elements and another group contains $n$ elements. We have to find number of ways to pick same number of elements from both the groups.




      My approach is
      $1+dbinom m1.dbinom n1+dbinom m2 . dbinom n2 +.... $upto $dbinom mm$ or $dbinom nn$ whatever is smaller.

      Is there any faster way to do the same?



      Example:
      $3$ and $2$
      $1+dbinom 31.dbinom 21+dbinom 32 . dbinom 22=10$ ways






      combinatorics permutations







      share|cite|improve this question









      New contributor




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












      There are two groups such that one group contains $m$ elements and another group contains $n$ elements. We have to find number of ways to pick same number of elements from both the groups.




      My approach is
      $1+dbinom m1.dbinom n1+dbinom m2 . dbinom n2 +.... $upto $dbinom mm$ or $dbinom nn$ whatever is smaller.

      Is there any faster way to do the same?



      Example:
      $3$ and $2$
      $1+dbinom 31.dbinom 21+dbinom 32 . dbinom 22=10$ ways






      combinatorics permutations




      combinatorics permutations






      share|cite|improve this question









      New contributor




      Sagar Sharma 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 question









      New contributor




      Sagar Sharma 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 question




      share|cite|improve this question








      edited yesterday









      SmarthBansal

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      36412






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      asked yesterday









      Sagar Sharma

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