Question

There are $(n+1)$ white and $(n+1)$ black balls, each set numbered from $1$ to $n+1$. The number of ways in which the balls can be arranged in a row so that the adjacent balls are of different colours is

• A. $(2n+2)!$
• B. $\left[\right(n+1)!{]}^2$
• C. $2\left[\right(n+1)!{]}^2$
• D. $(n+1)!$

Answer 1

The correct option is C $2\left[\right(n+1)!{]}^2$

Arranging all the white balls in $(n+1)!$ ways
Now there are $n+2$ gaps
Ways of arranging the black balls is $(n+1)!$
We can start with white or black ball
Therefore, the required number of arrangements
$2\times \left[\right(n+1)!\cdot (n+1)!]=2\left[\right(n+1)!{]}^2$