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,

$\mathrm{n}\left(\mathrm{S}\right)=5^5$

For 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 in

$3!\left[\frac{5!}{2!2!2!}+\frac{5!}{3!2!}\right]=150\text{ ways }$

$\Rightarrow \mathrm{n}\left(\mathrm{A}\right){=}^5{\mathrm{C}}_2\times 150$

Hence,

$\mathrm{P}\left(\mathrm{E}\right){=}^5{\mathrm{C}}_2\times \frac{150}{5^5}=\frac{60}{125}=\frac{12}{25}$