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?
 A: Let $A$ be the empty box, and $B$ be the box containing $2$ balls and $C$ is $n-2$ boxes containing $1$ ball. So, there are $2\dbinom{n}{2}$ ways of arranging the letter sequence $ABC.....C$. Then the size of the sample is $\dbinom{n+n-1}{n}$ because it is equivalent to the number of ways of placing $n$ $0$'s and $n-1$ $1$'s in the order. So the probability is $\dfrac{n(n-1)}{\dbinom{2n-1}{n}}$.
Can you now solve the problem?
A: Here is a solution by way of an extension of the Principle of Inclusion / Exclusion (PIE).
Say that an arrangement of balls in the boxes has "Property $i$" if box $i$ is empty, for $i=1,2,3,4$.  Let $S_j$ be the total of the probabilities of all the arrangements with $j$ of the properties, for $j=1,2,3,4$.  Then
$$\begin{align}
S_1 &= \binom{4}{1} \left( \frac{3}{4} \right)^5 \\
S_2 &= \binom{4}{2} \left( \frac{2}{4} \right)^5 \\
S_3 &= \binom{4}{3} \left( \frac{1}{4} \right)^5 \\
S_4 &= 0
\end{align}$$
An extension of PIE states that the probability that exactly $m$ of $N$ events will occur is
$$P_{[m]} = S_m - \binom{m+1}{m} S_{m+1} + \binom{m+2}{m} S_{m+2} - \dots \pm \binom{N}{m} S_N$$
Reference: An Introduction to Probability Theory and Its Applications, Volume I, Third Edition, by William Feller, Section IV.3.
Ours is the case $m=1$, $N=4$. So the probability that exactly one box is empty is 
$$P_{[1]} = S_1 - \binom{2}{1} S_2 + \binom{3}{1} S_3 - \binom{4}{1} S_4= \boxed{0.585938}$$
A: 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 $= 4$
$P(1 \text{empty}) = \frac{600}{240+600+180+4} = \frac{600}{1024} = .58594$
