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If the number of ways of choosing 2 boys and 2 girls in a class for a game of mixed doubles is 1620, what is the number of ways of choosing 2 students from the class?

My attempt: Let there be $m$ boys and $n$ girls. Number of ways of choosing 2 boys and 2 girls from them =$m \choose 2 $$ n \choose 2$. Now from each of these selections we can make 2 teams (If the boys are P & Q and the girls R & S, then the games will be P+R vs Q+S and P+S vs Q+R)

Therefore $2$$m \choose 2 $$ n \choose 2$= $1620$

How can I find $m+n \choose 2$?

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    $\begingroup$ Didn't you forget a factor $2$ ? If we choose $2$ girls and $2$ boys, we have still two possibilities to form the mixed doubles. $\endgroup$ – Peter Nov 21 '15 at 9:39
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    $\begingroup$ It should be $2\cdot (^m_2)(^n_2) = 1620$. $\endgroup$ – SchrodingersCat Nov 21 '15 at 9:40
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    $\begingroup$ But a solution only exists for $\binom{m}{2}\times \binom{n}{2}=1620$ (It is $m=9,n=10$ or vice versa) $\endgroup$ – Peter Nov 21 '15 at 9:43
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    $\begingroup$ Perhaps, the intention of the exercise is that the four children are a team, and it does not matter how to pair them. $\endgroup$ – Peter Nov 21 '15 at 9:45
  • $\begingroup$ @Aniket I forgot to write factor 2 in the expression but wrote it in the text. sorry $\endgroup$ – Archisman Panigrahi Nov 21 '15 at 12:56
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Hint: Do it like $2!.{m \choose 2}.{n \choose 2}=1620$. Now you can get answers by trial and error method.

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    $\begingroup$ And this method shows that there is no solution with the additional "$2$". $\endgroup$ – Peter Nov 21 '15 at 9:51
  • $\begingroup$ Thanks Aniket for editing. $\endgroup$ – Archis Welankar Nov 21 '15 at 10:05

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