The number of ways of arranging n students in a row such that no two b

1.	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 :1). 82). 123). 94). 10
Answer» If n is even, then the number of BOYS should be equal to number of girls, LET each be a So, 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 Hence, 3(2 × a! × a!) = a! × (a + 1)! Which gives (a + 1) = 6 and a = 5 As a result 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)! Hence, by the GIVEN condition, 2(a + 1) = 3 which is not possible.

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 :1). 82). 123). 94). 10

Answer»

If n is even, then the number of BOYS should be equal to number of girls, LET each be a

So, 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

Hence, 3(2 × a! × a!) = a! × (a + 1)!

Which gives (a + 1) = 6 and a = 5

As a result 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)!

Hence, by the GIVEN condition, 2(a + 1) = 3 which is not possible.

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