JEE Main · 2023 · 10 Apr 2023 · Shift 1 · Permutations and Combinations

Let there be $n$ couples, so $n$ men and $n$ women. A mixed-doubles match uses two men and two women with no married pair together. Choose the two men in $\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ ways; the two women must then come from the $n-2$ women not married to either chosen man, giving $\left(\genfrac{}{}{0ex}{}{n-2}{2}\right)$ ways; and the four players can be split into two mixed teams in $2$ ways. Hence $\left(\genfrac{}{}{0ex}{}{n}{2}\right)\left(\genfrac{}{}{0ex}{}{n-2}{2}\right)\cdot 2=840$, which simplifies to $n(n-1)(n-2)(n-3)=1680$. Since $8\cdot 7\cdot 6\cdot 5=1680$, we get $n=8$, so the number of participants is $2n=16$.