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 $= 10!.10!+10!.10!=2\times (10!)^2$