# Putting unlabelled balls into labeled boxes [duplicate]

How many ways are there to put $50$ unlabeled balls into $100$ labeled boxes? If empty boxes are allowed and you're also allowed to put as many balls as you want into a single box, would the answer be $\displaystyle\binom{149}{50}$?

## marked as duplicate by David K, Community♦Mar 4 '16 at 22:07

• Let $x_1, x_2,\dots,x_{100}$ be the number of balls in each box. It would be the number of non-negative integral solutions to $x_1+x_2+\dots+x_{100}=50$. What made you pick $149$? What made you pick $100$? – JMoravitz Mar 4 '16 at 21:57
• I actually did originally write 50. The 149 came from the formula $(n-1 + k) choose k$ – John Kyle Mar 4 '16 at 21:59
• You use /k. Remember the binomial coefficient $\binom{n}{r}$ is very different than $\frac{n}{r}$ and should not be confused. – JMoravitz Mar 4 '16 at 22:01
Let the balls be in line. Like this $$\circ \circ\circ\circ\circ\circ \dots \circ\circ\circ\circ\circ\circ$$ Then you divide them using $99$ barriers. Like this $$\circ \circ\circ\mid\mid \circ\ \circ\mid\circ \dots \circ\circ\mid\circ\circ\circ \mid \circ \mid \mid$$ The spaces between the barriers and the extreme left and extreme right spaces can be considered as (labelled, as they are ordered) boxes. Now, this becomes a permutation with repetition problem: you have $149$ elements, $50$ equal balls and $99$ equal barriers. So the answer is $$\dfrac{149!}{50!\cdot99!} = \binom{149}{50}$$