Question

In how many ways a mixed double games can be arranged from $8$ married couples if no husband and wife play in the same game?

Answer 1

2 men are selected from 8 men in ${}^8{C}_2$ ways since no husband and wife should be in same game.

2 woman out of remaining 6 are chosen in ${}^6{C}_2$ ways.

Now one team can be chosen as $(M_1,W_1)$ or $(M_1,W_2)$ in 2 ways.

$\therefore$ The required number of arrangements

${=}^8{C}_2{\times }^6{C}_2\times 2$

$\frac{8!}{6!2!}\times \frac{6!}{4!2!}\times 2$

$=\frac{8\times 7\times 6!}{6!\times 2!}\times \frac{6\times 5\times 4!}{4!\times 2!}\times 2$

$4\times 7\times 3\times 5\times 2$

= 840