$$7$$ white balls and $$3$$ black balls are placed in a row at random. The probability that no two black balls are adjacent is :

Solution :

$$n\left( S \right) = \frac{{10!}}{{\left( {7!} \right)\left( {3!} \right)}}$$
$$n\left( E \right) = {}^8{C_3} = \frac{{8!}}{{\left( {3!} \right)\left( {5!} \right)}},$$      because there are $$8$$ places for $$3$$ black balls.
$$\therefore \,P\left( E \right) = \frac{{\frac{{18!}}{{\left( {3!} \right)\left( {5!} \right)}}}}{{\frac{{10!}}{{\left( {7!} \right)\left( {3!} \right)}}}} = \frac{{\left( {8!} \right)\left( {7!} \right)}}{{\left( {10!} \right)\left( {5!} \right)}} = \frac{{7.6}}{{10.9}} = \frac{7}{{15}}.$$