Question

5 girls and 10 boys sit at random in a row having 15 chairs numbered as 1 to 15, then the probability that end seats are occupied by the girls and between any two girls an odd number of boys sit is

• A. $\frac{20\times 10!\times 5!}{15!}$
• B. $\frac{10\times 10!\times 5!}{15!}$
• C. $\frac{20\times 10!\times 30!}{15!}$
• D. $\frac{10\times 10!\times 5!}{25!}$

Answer 1

The correct option is A $\frac{20\times 10!\times 5!}{15!}$

The girls can be arranged in $\left(\frac{5!}{3!}\right)$ ways.

As both ends seats are occupied by $=20\times 5!$ ways

Girls and Boys are arranged addly between any $2$ girls.

The boys are occupying the odd places such that they can be arranged in $10!$ ways.

$\therefore$ The no. of ways of arrangement $=\frac{20\times 10!\times 5!}{15!}$

The total no. of chairs in the row can be arranged in $15!$ ways.

Hence, the answer is $\frac{20\times 10!\times 5!}{15!}.$