Question

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

• A. $840$
• B. $460$
• C. $580$
• D. none of the above

Answer 1

The correct option is A $840$

Given that there are $8$ couples,

So,the number of ways in which a mixed double game can be arranged such that no husband and wife play the same game are,

number of derangements of $8$ things,because if we assume the wife as gaps and the husbands as things,the thing should not go into its respective/own place,

$\therefore$ the number of ways of doing this are,

$8!(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+\frac{1}{6!}-\frac{1}{7!}+\frac{1}{8!})⟹8!(1-1+\frac{1}{2}-\frac{1}{6}+\frac{1}{24}-\frac{1}{120}+\frac{1}{720}-\frac{1}{7!}+\frac{1}{8!})⟹20160-6720+1680-336+56-8+1⟹14883$