# sir,i need an answer to this question from permutations and combinations.the question is:-there are n distinct white balls and n distinct black balls.if the number of ways of arranging them in a row such that neighbouring balls are of different colours is 1152 then value of n?

Akshay
185 Points
7 years ago
n=4,
suppose it starts with black ball. sequence will be BWBW......BWBW,
and if it starts with white ball, sequence will be WBWB.....WBWB.
so, there are two cases,
for each case, total will be = n! * n!, n! to permute the white balls and other n! to permute black balls.
so, 1152=2*n!*n!,
576=(n!)^2,
n!=24,
n=4
Vijay Mukati
7 years ago
Definately, you will have to start with white or black (2 cases) and then repeat the order as WBWBWB or BWBWBW … and ways to arrange n white or black balls will be factorial(n). Therefore 2*factorial(n)2 = 1152. Therefore n = 4.
manmath
49 Points
7 years ago
 n=4, suppose it starts with black ball. sequence will be BWBW......BWBW, and if it starts with white ball, sequence will be WBWB.....WBWB. so, there are two cases, for each case, total will ...