roopika2990 wrote:

Can anyone help me with this? I have no idea how to solve this.

Consider the possible arrangements:

WBWBWBWBWBWW or BWBWBWBWBWWW or WWBWBWBWBWBW or WWWBWBWBWBWB

Clearly, we are concerned about not placing 2 B's together, whites can be placed however we want.

So, it is like _W_W_W_W_W_W_W_ where black ball can be placed anywhere on the blank spaces, and this way, there never will be a case where 2 black balls fall together.

As there are 8 black spaces where 5 black balls can be placed, thus, the combination turns out as 8C5 (again, we are not concerned about the placement of white balls as we have already "fixed" their positions in the diagram above)

Also, we achieved at 8C5 as arranging 5 black balls on 8 different spots => 8!/5!3! (Total combination of available spaces = 8!; choosing 5 spaces out of them = 5!, arranging 3 leftover spaces = 3!) = 56

Hence answer is A