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 $m+n \choose 2$?