Five different marbles are placed in 5 different boxes randomly. Then the probability that exactly two boxes remain empty is (each box can hold any number of marbles)

We have,n(S)=55For computing favorable outcomes, 2 boxes which are to remain empty, can be selected in 5C2 ways and 5 marbles can be placed in the remaining 3 boxes in groups of 221 or 311 in3!5!2!2!2!+5!3!2!=150 ways ⇒ n(A)=5C2×150Hence,P(E)=5C2×15055=60125=1225