Seven white balls and three black balls are randomly placed in a...

White balls $=7$, black Balls $=3$, Total Balls $=10$ Total number of ways to place all these balls in a row $$\begin{aligned} & =\frac{10!}{7!\times 3!}=\frac{10 \times 9 \times 8 \times 7!}{7!\times 3 \times 2 \times 1} \\ & n=120 \end{aligned}$$ Now, The number of ways in which no two black balls put together is equal to the number of ways of choosing 3 places marked places of eight places. $-\mathrm{W}-\mathrm{W}-\mathrm{W}-\mathrm{W}-\mathrm{W}-\mathrm{W}-\mathrm{W}-$ This can be done in ${ }^8 \mathrm{C}_3 ways.$$\begin{aligned} & \therefore \text{ Required probability }=\frac{{ }^8 C_3}{120}=\frac{8!}{3!5!120} \\ & P=\frac{8 \times 7 \times 6 \times 5!}{3 \times 2 \times 5!\times 120}=\frac{7}{15} \end{aligned}$$$