There are 7 identical white balls and 3 identical blackballs. The number of distinguishable arrangements in a row of all the balls, so that no two black balls are adjacent is

First arrange 7 identical white balls in a row in $\frac{7!}{7!}=1$ way.

Now arrange 3 identical black balls in 8 places such that no two black balls are adjacent to each.. other is $\frac{{}^{8}P_3}{3!}$ ways

Required number of arrangements $=1\times \frac{{}^{8}P_3}{3!}=\frac{8\times 7\times 6}{6}=56$.