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) `2//5` b. `12//25` c. `3//5` d. none of these

Options

A. `(24)/(125)`
B. `(12)/(25)`
C. `(96)/(125)`
D. None of these

Answer

B

Explanation

`(b)` Each ball can be placed in 5 ways.
`:.` Total number of ways, `n(S)=5^(5)`
`2` empty boxes, can be selected in `"^(5)C_(2)` ways and `5` balls can be placed in the remaining `3` boxes in groups of `221` or `311` in
`3![(5!)/(2!2!2!)+(5!)/(3!2!)]=150` ways
`:.` Favorable cases `n(A)="^(5)C_(2)*150`
Hence `P(A)="^(5)C_(2)*(15)/(5^(5))=(60)/(125)=(12)/(25)`