Some couples participated in a mixed doubles badminton tournament....

Given,

Some couples participated in a mixed doubles badminton tournament. If the number of matches played, so that no couple played in a match, is $840$,

Now Let total number of persons be $2n$, so number of couples will be $n$

Now number of matches played, so that no couple played in a match is given by,

$⇒ C2 n \cdot C2 n - 2 \cdot 2 = 840$

{as if we choose couple as \left\{A, B\right\}$and other couple as \left\{C, D\right\}$ and taking other two couple as \left\{x, y\right\} \text{ and } \left\{w, z\right\}, so the possible combination of the match where no couple played in a match will be, \left\{A, x\right\}$\to \left\{C, w\right\}\text{\text{ and } }$ \left\{C, x\right\}$\to \left\{A, w\right\}$, hence there are two possible combination after choosing }

$⇒ n\left(n - 1\right)\left(n - 2\right)\left(n - 3\right) = 5 \cdot 6 \cdot 7 \cdot 8$

So, $n = 8$

Hence, the number of persons will be, $2n = 16$