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?
Best Answer
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?