Question

5 girls and 10 boys sit at random in a row having 15 chairs numbered as 1 to 15. If the probability that the end seats are occupied by the girls and odd number of boys take seat between any two girls is $\frac{20}{n}$, then the value of $\frac{3003}{n}$ is

Answer 1

Total number of arrangements = 15!

Extreme chairs are occupied by girls, thus there are four gaps among 5 girls where boys can be seated. Let the number of boys in these four gaps be $2x+1,2y+1,2z+1$ and $2t+1$, then

$2x+1+2y+1+2z+1+2t+1=10$

$\Rightarrow x+y+z+t=3$

Where $x,y,z,t$ are integers and $0\le x\le 3,0\le y\le 3,0\le z\le 3,0\le t\le 3$

$\therefore$ The number of ways of selecting positions for boys = coefficient of $x^3$ in $(1+x+x^2+x^3{)}^4$

= coefficient of $x^3$ in ${\left(\frac{1-x^4}{1-x}\right)}^4$

= coefficient of $x^3$ in $(1-x^4{)}^4(1-x{)}^{-4}={}^6{C}_3=20$

$\therefore$ Number of arrangements of boys and girls with given condition is $20\times 10!\times 5!$

$\therefore$ Required probability $=\frac{20\times 10!\times 5!}{15!}=\frac{20}{3003}\Rightarrow n=3003$

$\Rightarrow$ The value of $\frac{3003}{n}$ is 1