Question

There are $(\mathrm{n}+1)$ white and $(\mathrm{n}+1)$ black balls each set numbered $1$ to $(\mathrm{n}+1)$. The number of ways in which the balls can be arranged in a row so that the adjacent balls are of different colors is ?

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

Answer 1

The correct option is D

$2{\left[\left(\mathrm{n}+1\right)!\right]}^2$

Evaluate the required number of ways

The white and black balls can be arranged in two ways:

$\mathrm{BWBWBWBW}$ or $\mathrm{WBWBWBWB}$

Now, if we take the first arrangement, then $\left(\mathrm{n}+1\right)$ black balls can be arranged in $(\mathrm{n}+1)!$ Ways, where the $(\mathrm{n}+1)$ white balls will also get arranged in $(\mathrm{n}+1)!$ ways

Thus, total number of arrangements $=\left(\mathrm{n}+1\right)!\times \left(\mathrm{n}+1\right)!$

$=(\mathrm{n}+1){!}^2$

Similarly, if we take the second arrangement, then we get ${(\mathrm{n}+1)!}^2$ ways.

Therefore, total number of arrangements$=\left(\mathrm{n}+1\right){!}^2+{(\mathrm{n}+1)!}^2$

$=2{\left[(\mathrm{n}+1)!\right]}^2$

Hence option $\left(\mathrm{D}\right)$, $2{\left[\left(\mathrm{n}+1\right)!\right]}^2$ is the correct answer.