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. $20!\times {}^{21}{P}_{19}$
• B. $20!\times 19!$
• C. $(20!{)}^2$
• D. $\left(21\right)!\times {}^{20}{C}_{19}$

Answer 1

The correct option is B $20!\times 19!$

Let $20$ white balls be $W_1,W_2,W_3,...,W_{20}$
and $19$ black balls be $B_1,B_2,...,B_{19}.$
Since no two balls of same colour come together, balls should alter colour successively.
So,
$\left(1\right)$ If arrangement starts with Black coloured ball, it is not possible, since $2$ white coloured balls come together
$\left(2\right)$ If arrangement starts with white coloured ball then specific arrangement can be
$W_1{B}_1{W}_2{B}_2...W_{19}{B}_{19}{W}_{20}$
Here White balls can be arranged between themselves in $20!$ ways.
and
Black balls can be arranged between themselves in $19!$ ways
$\therefore$ Required number of ways $=20!19!$