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 and odd number of boy sit is $5\mathrm{K}*10!*5!/15!$. Find $\mathrm{K}$

Answer 1

There are four gaps in between the girls where the boys can sit.

Step $1$

Let the number of boys in these gaps be

$2\mathrm{a}+1,2\mathrm{b}+1,2\mathrm{c}+1,2\mathrm{d}+1$

Then,

$$\begin{gathered} 2\mathrm{a}+1+2\mathrm{b}+1+2\mathrm{c}+1+2\mathrm{d}+1=10 \\ \Rightarrow \mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}=3 \end{gathered}$$

Step $2$

The number of solution of above question=cofficient of ${\mathrm{x}}^3\mathrm{in}{\left(1-\mathrm{x}\right)}^4=6_{{\mathrm{C}}_3}=20$

Here both the boys and girls can sit in$20*10!*5!$ways

Therefore, total ways$=15!$

The required probability $=(20*10!*5!)/15!$

$\mathrm{k}=4$

Hence, $\mathrm{k}=4$