The number of ways of arranging n students in a row such that no two boys sit together and no two girls sit together is m(m > 100). If one more student is added, then number of ways of arranging as above increases by 200%. The value of n is:

Solution (By Examveda Team)

If n is even, then the number of boys should be equal to number of girls, let each be a.
⇒ n = 2a
Then the number of arrangements = 2 × a! × a!
If one more students is added, then number of arrangements,
= a! × (a + 1)!
But this is 200% more than the earlier
⇒ 3 × (2 × a! × a!) = a! × (a + 1)!
⇒ a + 1 = 6 and a = 5
⇒ n = 10
But if n is odd, then number of arrangements, = a!(a + 1)!
Where, n = 2a + 1
When one student is included, number of arrangements,
= 2(a + 1)! (a + 1)!
By the given condition, 2(a + 1) = 3, which is not possible.