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 colour, is

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

Answer 1

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

Let the $10$ white balls be arranged in $10!$ ways in an alternate manner, having a space of one ball in between
Now, the $10$ black balls can be arranged in $10!$ ways, either starting as first ball or next to black ball from second place.
Total ways $=10!\times 10!+10!\times 10!=2(10!{)}^2$

Alternate method :
Let $10$ white balls be $W_1,W_2,W_3,...,W_{10}$
and $10$ black balls be $B_1,B_2,...,B_{10}.$
Since no two balls of same colour come together, balls should alter colour successively.
So,
$\left(1\right)$ If arrangement starts with white coloured ball, then specific arrangement can be
$W_1{B}_1{W}_2{B}_2...W_9{B}_9{W}_{10}{B}_{10}$
Here white balls can be arranged between themselves in $10!$ ways.
and
Black balls can be arranged between themselves in $10!$ ways
$\therefore$ Number of ways for such arrangement $=10!10!$
$\left(2\right)$ If arrangement starts with black coloured ball then specific arrangement can be
$B_1{W}_1{B}_2{W}_2...B_9{W}_9{B}_{10}{W}_{10}$
Here white balls can be arranged between themselves in $10!$ ways.
and
Black balls can be arranged between themselves in $10!$ ways
$\therefore$ Number of ways for such arrangement $=10!10!$
Hence, Required Number of ways $=$ case$1$ $+$ case$2$
$=10!10!+10!0!$
$=2\times 10!10!$
$=2(10!{)}^2$