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 $=\frac{5k\times 10!\times 5!}{15!}$.Find $k$

Answer 1

There are four gaps in between the girls where the boys can sit.
Let the number of boys in these gaps be $2a+1,2b+1,2c+1,2d+1$
Then $2a+1+2b+1+2c+1+2d+1=10\Rightarrow a+b+c+d=3$
The number of solution of above equation $=$ coefficient of $x^3$ in ${(1-x)}^4{=}^6{C}_3=20$
These boys and girls can sit in $20\times 10!\times 5!$ ways
Total ways $=15!$
Hence the required probability $=\frac{20\times 10!\times 5!}{15!}$
$\therefore k=4$