Some couples participated in a mixed doubles badminton tournament. If the number of matches played, so that…

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,

$\Rightarrow {}^{n}C_2·{}^{n-2}C_2·2=840$

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

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

So, $n=8$

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