Question

There are $10$ white and $10$ black balls marked $1,2,3.....10.$ The number of ways in which we can arrange these balls in a row, in such a way that, neighbouring balls are of different colours is

• A. $10!9!$
• B. $20!$
• C. $(10!{)}^2$
• D. $2(10!{)}^2$

Answer 1

The correct option is D $2(10!{)}^2$

$\to$ The black and white balls have to be placed alternately. However, the first ball may be black or white.
$\to$ So,
Case I $\to 1^{st}$ ball is black.
$C_I=10!\times 10!$
Case II $\to 1^{st}$ ball is white.
$C_{II}=10!\times 10!$
$\to$ So, the total no.$=10!\times 10!+10!\times 10!$
$\Rightarrow Ans=2(10!{)}^2$