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

n(S)=(10−1)!=9! n(E)=8!×2! [2 Boys−1 unit8 Boys−8 units] The probability of arranging them so that two specified boys of those ten must come together is=P(E)=8!×2!9!=29