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So assume there are $N$ identical balls to be arranged into $r$ boxes. There could be empty boxes. Howe many ways are there to arrange it?

So the text book give an example of solving $N$ identical balls into $r$ boxes without empty boxes. as such: $$ x_1 + x_2 + \cdots + x_r = N, \quad x_i > 0 $$ and the solution to that is simply $N-1 \choose r-1$. And I can interpret that as if choose $r-1$ spaces from $N-1$ spaces between balls.

but then the text book solve the empty boxes case: $$ x_1 + x_2 + \cdots + x_r = N,\quad x_i \geq 0 $$

by substituting $ y_i - 1 = x_i $ so that now $ y_i > 0. $

now the questions go back to the first situation. so the solution is $ N+r-1 \choose r-1$

But I can't seem to interpret the results using combinatorial arguments. What is the best way to interpret this result?

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up vote 1 down vote accepted

Put $N+r-1$ objects in a line, and select $r-1$ of them, and call them dividers, dividing the objects (if any) in one box from the objects (if any) in the next.

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