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

Answer 1

The correct option is A

$60\times 5!$

The four people, i.e. A, B and the two persons between them are always together. Thus, they can be considered as a single person.

So, along with the remaining 4 persons, there are now total 5 people who need to be arranged,This can be done in 5! ways.

But, the two persons that have to be included between A and B could be selected out of the remaining 6 people in ${}^6{P}_2$ ways, which is equal to 30.

For each selection, these two persons standing between A and B can be arranged among themselves in 2 ways.

$\therefore$ Total number of arrangements = $5!\times 30\times 2=60\times 5!$