Evaluating $\sum_{i=0}^{n-1}\sum_{j=i+1}^{n-1}\sum_{k=j+1}^{n-1}1$ 
How can I evaluate this triple sum?
$$\large\sum_{i=0}^{n-1}\sum_{j=i+1}^{n-1}\sum_{k=j+1}^{n-1}1$$

I started to calculate but I struggled in the second internal sum.
Update

This is the innermost sum that I tried:
$$\large\sum_{i=0}^{n-1}\sum_{j=i+1}^{n-1}(n-j-1)$$
$$\large\sum_{j=i+1}^{n-1}(n-j-1)=\sum_{j=i+1}^{n-1}n - \sum_{j=i+1}^{n-1}j-\sum_{j=i+1}^{n-1}1$$
$$\large\sum_{j=i+1}^{n-1}n= n^2 - n - ni$$
 A: The way to do this is to evaluate the sums iteratively, working from the inner most to the outer most sum. As we go along, we will need to use the following three identities
$$
\underbrace{\sum_{i=1}^{m}1 = m}_{\textbf{(A)}},\quad \underbrace{\sum_{i=1}^{m}i^2 = \frac{1}{2}m(m+1)}_{\textbf{(B)}}, \quad \underbrace{\sum_{i=1}^{m}i^2 = \frac{1}{6}m(m+1)(2m+1)}_{\textbf{(C)}}, 
$$
for some arbitrary positive integer $m$.
Having laid out the key expressions, we can begin this in the following way.
Inner Most Sum
Firstly, note that we can rewrite the inner most sum in such a way that we can apply (A), and indeed doing so, we obtain
$$
\sum_{k=j+1}^{n-1}1 = \sum_{k = 1}^{n-1-j}1 = n-1-j
$$
Middle Sum
Having evaluated the inner most sum, we can proceed and obtain the following
$$
\begin{align}
\sum_{j=i+1}^{n-1}\sum_{k=j+1}^{n-1}1 &= \sum_{j=i+1}^{n-1}(n-1-j)\\
\text{separate the sum gives }&= \sum_{j=i+1}^{n-1}(n-1) - \sum_{j=i+1}^{n-1}j\\
\text{use (A) and (B)}&= (n-1)(n-1-i) - \frac{1}{2}(n-1-i)(n+i)\\
&= \frac{1}{2}(n-1-i)(n - 2 - i) \\
&= \frac{1}{2}\left[(n-1)(n-2) - i(n-1) - i(n-2) + i^2\right] \\
&= \frac{1}{2}\left[(n-1)(n-2) - i(2n - 3) + i^2\right].
\end{align}
$$
Outermost Sum
We are now left with the final sum.
$$
\begin{align}
\sum_{i=0}^{n-1}\sum_{j=i+1}^{n-1}\sum_{k=j+1}^{n-1}1 
&= \frac{1}{2}\sum_{i=0}^{n-1}\left[(n-1)(n-2) - i(2n - 3) + i^2\right] \\
\text{separate the sum}
&= \frac{1}{2}\left[n(n-1)(n-2) - (2n-3)\sum_{i=0}^{n-1}i + \sum_{i=0}^{n-1}i^2\right] \\
\text{use (A), (B) and (C)}
&= \frac{1}{2}\left[n(n-1)(n-2) - \frac{1}{2}n(n-1)(2n-3) + \frac{1}{6}n(n-1)(2n-1)\right]
\end{align}
$$
Simplifying down, we find that
\begin{align}
&\frac{1}{2}\left[n(n-1)(n-2) - \frac{1}{2}n(n-1)(2n-3) + \frac{1}{6}n(n-1)(2n-1)\right]\\
&= \frac{1}{2}n(n-1)\left[n-2 - \frac{1}{2}(2n-3) + \frac{1}{6}(2n-1)\right] \\
&= \frac{1}{6}n(n-1)(n-2). 
\end{align}
Final Result
Thus, we can conclude that
$$
\sum_{i=0}^{n-1}\sum_{j=i+1}^{n-1}\sum_{k=j+1}^{n-1}1 = \frac{1}{6}n(n-1)(n-2).
$$

Extension Result
The above exercise leads to the following interesting exercise. Given some index $m$, what is the evaluation of the sum (whenever it makes sense)
$$
\sum_{i_1=0}^{n-1}\sum_{i_2=i_1+1}^{n-1}\sum_{i_3=i_2+1}^{n-1}\cdots\sum_{i_m=i_{m-1}+1}^{n-1}1 ?
$$
This is quite an interesting question, and yields a very nice answer. Indeed, we have that
$$
\sum_{i_1=0}^{n-1}\sum_{i_2=i_1+1}^{n-1}\sum_{i_3=i_2+1}^{n-1}\cdots\sum_{i_m=i_{m-1}+1}^{n-1}1 = \frac{1}{m!}\prod_{i=0}^{m-1}(n-i) = \frac{n!}{m!(n-m)!} = {n \choose m}
$$
You can find out more about sums of this form in this article.
A: The sum is equal to the number of ordered triples of integers $(i, j, k)$ such that $0 \leqslant i < j < k \leqslant n - 1.$ Therefore, it is equal to the number of $3$-element subsets of the $n$-element set $\{0, 1, \ldots, n - 1\},$ which is $\binom{n}3.$ Similarly for the more general result given in Spaceman's answer.
