Question

If a couple dance competition happens between $7$ married couples, then the number of such possible pairs that can be made such that no couple dances in the same pair is

• A. $420$
• B. $1540$
• C. $1854$
• D. $2520$

Answer 1

The correct option is C $1854$

Here $7$ men has be paired with $7$ women such that no one pairs with his own wife . So required number of ways is
$=7!(\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+\frac{1}{6!}-\frac{1}{7!})=1854$