This question appeared as an unsolved exercise in an introductory combinatorics textbook:
We have a bag with as many identical balls as necessary. We take out i balls and put them in 10 numbered boxes, such that the number of balls in box 1 $\leq $ the balls in box 2 $\leq $ … $\leq $ the number of balls in box 10 $\leq $ 20. How many ways are there to do this?
A hint is provided (sorry for the shoddy translation): "Think of the sum of the series of differences of the amount of balls in adjacent boxes. The first difference equals the amount of balls in box 1".
Background: Earlier today I asked this question: Number of integer solutions if $x_1\leq x_2\leq x_3\leq \cdots\leq x_r\leq k$. The question itself was based on this exercise, so I was quite baffled to receive relatively complex answers. I figured I'll ask the question as it appeared, to see if maybe my 'generalisation' made this more complex than it should've been by omitting details.
Thank you for your help as always.
Best Answer
You didn't make this more complex by omitting details, but by adding them. In that other question, you had a fixed sum $k$. The present question doesn't place any requirements on the sum, and that makes it a lot easier. As the hint provided indicates, there's a bijection between the configurations to be counted and the ways of distributing $20$ balls into $11$ boxes: If we denote the number of balls in box $j$ by $n_j$ and add two extra numbers $n_0=0$ and $n_{11}=20$, the $11$ differences between these numbers add up to $20$, and each such sequence of differences determines one of the desired configurations and vice versa. So the desired number is
$$\binom{20+11-1}{11-1}=\binom{30}{10}=30045015\;.$$