Question

$4$ boys and $2$ girls occupy seats in a row at random. Then the probability that the two girls occupy seats side by side is

• A. $\frac{1}{2}$
• B. $\frac{1}{4}$
• C. $\frac{1}{3}$
• D. $\frac{1}{6}$

Answer 1

Answer: (C)

$\frac{1}{3}$

Girls sit side by side, it means both girls are seated together in $2$! ways
Let $n\left(E\right)=4$ boys and $2$ girls occupy seats side by side
$=5!2!$
$n\left(S\right)=6$!
$\therefore$ Required probability$=\frac{5!2!}{6!}=\frac{2}{6}=\frac{1}{3}$