Question

The number of ways in which $20$ different white balls and $19$ different black balls be arranged in a row. So that no two balls of the same colour come together is

• A. $20!{}^2{}^{1}P{}_1{}_9$
• B. $20!\times 19!$
• C. $(20!{)}^2$
• D. $\left(21\right)!{}^2{}^{0}C{}_1{}_9$

Answer 1

Answer: (B)

$20!\times 19!$

$\to$ This can only happen if the first ball is white and all the balls that follow are alternately black and white.
$\to$ So, the total no. of spaces available for black bolls $=19$. Hence, total ways of rearranging them $=19!$.
$\to$ Also, the total no. of spaces available for white bolls is $20.$ So, for white, no. of ways $=20!$
So, total no. $=20!\times 19!$
$\Rightarrow =20!\times 19!$