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,
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) 15 × 4! × 5!

(c) 4! × 5!

(d) none of these.

Solution:

(a) 60 × 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 6P2 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 $=5 ! \times 30 \times 2=60 \times 5 !$