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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

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  • $\begingroup$ 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 $100$? $\endgroup$ – JMoravitz Mar 4 '16 at 21:57
  • $\begingroup$ I actually did originally write 50. The 149 came from the formula $(n-1 + k) choose k$ $\endgroup$ – John Kyle Mar 4 '16 at 21:59
  • $\begingroup$ You use /k. Remember the binomial coefficient $\binom{n}{r}$ is very different than $\frac{n}{r}$ and should not be confused. $\endgroup$ – 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 $$\dfrac{149!}{50!\cdot99!} = \binom{149}{50}$$


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