Question

The number of different ways in which $8$ persons can stand in a row so that between two particular persons $A$ and $B$ there are always two persons, is

• A. $60(5!)$
• B. $4!\times 5!$
• C. $15(4!)\times 5!$
• D. None of these

Answer 1

The correct option is A $60(5!)$

The number of selection of $4$ persons including $A,B$

Considering these four as a group, the number ${=}^6{C}_2$ of arrangements with the other four $=5!$

But in each group of four, number of arrangements $=2!\times 2!$

$\therefore$ The required number of ways ${=}^6{C}_2\times 5!\times 2!\times 2!$.