What is the probability that exactly one box remains empty Five identical(indistinguishable) balls are to be randomly  distributed into $4$ distinct boxes. What is the probability that exactly one box remains empty?
we can choose the empty box in $4$ ways.
We should fit the $5$ balls now into $3$ distinct boxes in $^7C_5$ ways.
The correct answer must be: $\frac37$ (mcq)
Can you tell me how should I proceed then?

I couldn't see how the previous answer came up with a solution so I did it differently.
$0$ boxes empty has $1,1,1,2$ box contents times $\frac{5!}{2!}$ ball arrangements $= \frac{4!}{3!}\cdot \frac{5!}{2!} = 240$
$1$ box empty has $1,1,3$ and $1,2,2$ box contents and $\frac{5!}{3!}$ and $\frac{5!}{2!\cdot 2!}$ ball arrangements $= (\frac{3!}{2!}\cdot \frac{5!}{3!}+\frac{3!}{2!}\cdot \frac{5!}{2!\cdot 2!})\cdot ^4C_1 = 600$
$2$ boxes empty has $1,4$ and $2,3$ box contents and $\frac{5!}{4!}$ and $\frac{5!}{3!\cdot 2!}$ ball arrangements $=(2!\cdot \frac{5!}{4!}+2!\cdot \frac{5!}{3!\cdot 2!})\cdot ^4C_2 = 180$
$3$ boxes empty has all $5$ in any one of $4$ boxes $= 4$
$P(1 \text{empty}) = \frac{600}{240+600+180+4} = \frac{600}{1024} = .58594$
