combinatorics - Number of r-tuples with integer entries in a given interval

This is probably an easy question, but I couldn't figure it out.

Let $n$ and $r$ be positive integers with $n \geq r$. Then the number of elements of the set
$$\begin{align} \{(a_1,a_2,...,a_r) : 0 \leq a_1 \leq a_2\leq ... \leq a_r \leq n,\,\, a_i \in \mathbb{Z}\} \end{align}$$

is equal to the following $r$-fold sum,
$$\begin{align} \sum\limits_{a_1=0}^{n} \sum\limits_{a_2=a_1}^{n} .... \sum\limits_{a_r=a_{r-1}}^{n} 1 \end{align}$$

My question: Is there any simple expression for the above sum in terms of some binomial terms?

Answer

We are looking for number of tuples
$$\begin{align} \{(a_1,a_2,...,a_r) : 0 \leq a_1 \leq a_2\leq ... \leq a_r \leq n,\,\, a_i \in \mathbb{Z}\} . \end{align}$$

Let us say that you take random $r$ elements (you allow repetitions) from set: $$\left\{0,1,2,...,n \right\}$$
and you got
$$a_1,a_2,...,a_r$$
In how many ways can you put them to the tuple such that $0 \leq a_1 \leq a_2\leq ... \leq a_r \leq n$?

It is exactly $1$ way. So you are looking for the just number of possible ways to take $r$ elements from $\left\{0,1,2,...,n \right\}$.

From stars and bars you can do that in exactly
$$\binom{(n+1) + r - 1}{r}$$
I put $n+1$ instead of $n$ because you have $n+1$ elements in $\left\{0,1,2,...,n \right\}$

Example

$n=2, r=3$
$$2,2,2\\ 1,2,2\\ 1,1,2\\ 1,1,1\\ 0,1,1\\ 0,0,1\\ 0,0,0\\ 0,1,2\\ 0,0,2\\ 0,2,2$$
and it is equal to $\binom{(2+1)+3-1}{3} = 10$