Is it Variation? Counting elements Lets assume that we have a element, which can have value from 1 till n. (let's set it on 20 to make it easier)
And we have the Set, that consists of object, which consists of three elements $\langle e_1, e_2, e_3 \rangle$.
We have also one rule regarding to objects in the set: $e_1 \geq e_2 \geq e_3$
- example of good objects: $\langle n, n, n\rangle$, $\langle n, n-1, n-1\rangle$, $\langle 20, 19, 18\rangle$, $\langle 3, 2, 1\rangle$, $\langle 3, 3, 3\rangle$, $\langle 3, 2, 2\rangle$.
- example of bad objects: $\langle n, n+1, n\rangle$, $\langle 2, 3, 2\rangle$, $\langle 3, 2, 4\rangle$.  
Now the question:
How to count the amount of all good objects, which pass to this Set (they don't violate the rule ) ?
Can you give me any hints?
I can solve this with brute force method. But probably there is a short way.

If the first number is $k$, and the second number is $j$, where $j \leq k$ then the last number has $j$ choices.
So the number of favorable cases is $$\sum_{k=1}^n \sum_{j=1}^k  j = \sum_{k=1}^n \frac{k(k+1)}{2} = \frac{n(n+1)(n+2)}{6}$$
In general, if you have elements from $1$ to $n$ and want to choose an $m$ element set with the ordering you want the answer is $$\binom{n+m-1}{m}$$ which can be seen by induction on $m$ or by a combinatorial argument as proved here.
