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.

Answer 1

(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.
∴ Total number of arrangements = 5!$\times$30$\times$2 = 60$\times$5!