Question : Five different games are to be distributed among 4 children randomly. The probability that each child get atleast one game is

(A) 1/4

(B) 15/64

(C) 21/64

(D) none of these

Answer : option (B)

Solution :

The total number of ways of distribution is 4^{5}

therefore, n(S) = 4^{5}

The total number of ways of distribution so that each child gets at least one game is

4^{5}-4C_{1}3^{5}+4C_{2}2^{5}-4C_{3}

therefore, n(E) = 240,

Hence, the required probability,

\dfrac{n\left( E\right) }{n\left( S\right) }=\dfrac{240}{4^{5}}=\dfrac{15}{64}