N O bo O O LO LO O O O O LO O N N LO

N O bo O O LO LO O O O O LO O N N LO 5. There are 8 married couples and a mixed table-tennis game is to be arranged so that a and a wife do not play in the same game. How many games are possible?

Dear Student,
Out of total 8 men, lets suppose two men (M_{1 ,} M₂₎​​ chosen to play which is = ⁸C_{2 = $\frac{8!}{2! 6!}$ = 28}
We know that wives of two males already chosen cannot play. Total females left = 6
Out of 6 females 2 (F_{1 ,} F₂) are selected = ⁶C₂ = $\frac{6!}{2! 4!}$ = 15
Also F₁ can be arranged with two males in two ways
Total Games = 28 x 15 x 2 = 840

Regards