Ten boys are arranged at random along a circle. The probability of arranging them so that two specified boys of those ten must come together is

Ten boys are arranged at random along a circle

$n\left(S\right)=9!$

E= Event of of arranging them so that two specified boys of those ten must come together is

$n\left(E\right)=\left(9-1\right)!\times 2!=8!\times 2!$

$P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{8!\times 2!}{9!}=\frac{2}{9}$

$\text{ If }\mathrm{n}\text{ men among whom }\mathrm{A}\text{ and }\mathrm{B}\text{ sit along a circle then the probability that A,B sit together is }\frac{2}{n-1}$