Question

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

• A. $\frac{20\times 10!\times 5!}{15!}$
• B. $\frac{20\times 10!}{15!}$
• C. $\frac{20\times 5!}{15!}$
• D. None of these

Answer 1

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

There are four gaps in between the girls where the boys can sit.
Let the number of boys in these gaps be $2a+1,2b+1,2c+1,2d+1=0$.

Then $2a+1+2b+1+2c+1+2d+1=10\Rightarrow a+b+c+d=3$

The number of solutions of above equation $=$ coefficients of $x^3$ in ${(1-x)}^{-4}{=}^6{C}_3=20$

Thus, boys and girls can sit in $20\times 10!\times 5!$ ways

Total ways $=15!$

Hence, the required probability $=\frac{20\times 10!\times 5!}{15!}$