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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.

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up vote 1 down vote accepted

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.

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There are many ways. One is first to ask how many 2-element things there are. If $e_1=i$, there are $n+1-i$ of them. So the total number of 2-element things is $$\sum_{i=1}^n(n+1-i)=n^2+n-\frac{n(n+1)}{2}=\frac{n(n+1)}{2}$$
Then the number of 3-element things is $$\sum_{i=1}^n\frac{i(i+1)}{2}$$

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The simplest way is to simply pick three objects from your set allowing repetitions. Then you'll just list them in nonincreasing order.

This is just a count of combinations with repetitions. You have $n$ possibilities, and you want to pick $r$ elements, then the number of ways of doing it is $\binom{n+r-1}{r}$. See for example Wikipedia's page on the stars and bars problem for a proof.

In your example, $n=20$, $r=3$, so you would want $$\binom{20+3-1}{3} = \binom{22}{3} = 1540.$$

This agrees with Ross's answer: $$\sum_{i=1}^{20}\frac{i(i+1)}{2} = 1540.$$

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