Question

$7$ boys and $3$ girls are seated in a row randomly. The probability that no boy sit between two girls is

• A. $\frac{7!\times 3!}{10!}$
• B. $\frac{1}{15}$
• C. $\frac{8!}{10!3!}$
• D. $\frac{8!}{10!7!}$

Answer 1

Answer: (B)

$\frac{1}{15}$

$n\left(s\right)=10!$
$n\left(A\right)=$ the number of arrangement of $7$ boys and $3$ girls when all girls are together $=8!\times 3!$
$\therefore$ Required probability $=\frac{n\left(A\right)}{n\left(S\right)}$
$=\frac{8!\times 3!}{10!}=\frac{6}{10\times 9}=\frac{1}{15}$