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}