Question

There are ten persons among whom two are brothers. The total number of ways in which these persons can be seated around a table so that exactly one person sit between the brothers, is

• A. $2!\times 7!$
• B. $2!\times 8!$
• C. $3!\times 7!$
• D. none of these

Answer 1

The correct option is A $2!\times 7!$

Total Persons $=10$

Let a person be selected to sit between brothers ${=}^8{C}_1$

Let brothers and the selected person be a set.

Total arrangements around the circle$=(n-1)!=7!$

The brothers can be interchanged in $2!$ Ways

Therefore total arrangements $=7!\times 2!$