There are 10 person among whom two are brother. The total number of ways in which these persons can be seated around a round table so that exactly one person sit between the brothers , is equal to:

Correct answer is 8!*2!

We have 10 people and we can arrange them in 9! ways in order to identify the start point of cycle ( round table). For the rest of 9 people, we must take into account that exactly one person sits between the two brothers. 3 people out of 9 people must be considered as one unit and turn out, we get
{[(9-3)+1]=7}. Furthermore, we can arrange the two brothers in 2! ways and the answer is
7!*2!.
#Circular Permutation

Yes 2!*8! is the right answer.
because let b1,and b2 are brother.
so there are 8 possible ways to fix one person between them and brother can change
their positions in 2 ways :- 8*2!
now let b1pb2 is one person and remaining seven person.we will have to arranged in circle.
so total number of way to arranged is
=8*2!*(8-1)! (because arrangement in circle of n person is(n-1)! if clockwise and anticlockwise is not symmetric)
=2!*8!
b1pb2 and remaining 7 person .

fix one person in b/w B1 P B2
Brothers can be arranged 2!
remaining 7members can be arranged in 7!
but in between brothers B1 __ B2 can be filled by 8 person
so, 2!*7!*8=2!8!

The correct answer is 2! × 8!

Is the answer “b”?

option b is the right answer.

n will be n-1 because it's a circle that we shouldn't change the place of on of them to identify the start one of the circle.
then n-1-2 to make the three persons next to each other (two brothers and the one between them) so it will be 10-1-2=7 so 7!
and then we won't change the place of the one between two brothers but we will change of the place of the two brothers so it will be 2!
so 7!*2!

fix one person in b/w B1 P B2
Brothers can be arranged 2!
remaining 7members can be arranged in 7!
but in between brothers B1 __ B2 can be filled by 8 person
so, 2!*7!*8=2!8!

The answer is B)2!*8!

ans is B) 2!*8!