Combi Offer #6

Number of ways to pick 2 women

$\times$ number of ways to pick 2 men

$\times$ number of (MW-MW)-combinations

$= {4\choose 2}\times {6\choose 2}\times 2$

$=6\times 15 \times 2$

$=\boxed{180}.$

We have $10$ players which have $6$ men and $4$ women...

So ,according to condition ,a game is being played with mixed doubles ,i.e. a pair of players from one end and the other pair of players from the other end, i.e. $2+2=4$ players.

So ,number of combinations possible between $10$ players consisting of $6$ men and $4$ women is :-

$2(^6C_{2} . ^4C_{2})$

= $2(\dfrac{6!}{2! .4!} . \dfrac{4!}{2! . 2!})$

= $2(\dfrac{6 . 5}{2 . 1} . \dfrac{4 . 3}{2 . 1} )$

= $\color{#3D99F6}{180}$