# There are $10$ white and $10$ black balls each set numbered 1,2,3......10. The no. of ways in which they can be arranged so that no two adjacent balls are of same colour is ?

Case (i)
First let us place 10 white balls in (10!) ways.
Then 10 black balls are to be placed in between the white balls in (10!) ways.
Case (ii)
First place 10 black balls in 10! ways and
then place the white balls in between the black balls in 10! ways.
$\therefore$ The required no. of arrangements $= 10!.10!+10!.10!=2\times (10!)^2$
answered Sep 5, 2013