Seven white balls and three black balls are randomly arranged in a row. The probability that no two black…
Seven white balls and three black balls are randomly arranged in a row. The probability that no two black balls are placed adjacently is
$\frac{1}{2}$
$\frac{7}{15}$
$\frac{2}{15}$
$\frac{1}{3}$
Solution
Firstly we fix the seven white balls in alternate position in 7 !. In out of 8 positions 3 black balls can be placed in ${ }^8 P_3$ ways.
$\begin{aligned}
& \therefore \text { Required probability }=\frac{7 ! \times{ }^8 P_3}{10 !} \\
& =\frac{7 ! \times 8 !}{5 ! \times 10 !} \\
& =\frac{8 \times 7 \times 6}{10 \times 9 \times 8}=\frac{7}{15}
\end{aligned}$