TITLE: Finding sum : ${\mathop{\sum\sum\sum\sum}_{0\le i\lt j\lt k\lt l\le n }} \,1$
QUESTION [8 upvotes]: Finding the value of: $${\mathop{\sum\sum\sum\sum}_{0\le i\lt j\lt k\lt l\le n }} 1$$
I know a similar question was asked on this site earlier, but I couldn't understand the method used there.
Link to similar question
The answer of this question is $\binom{n+1}{4}$.
My try:
I tried tackling a smaller problem with two sums:
$$\mathop{\sum\sum}_{0\le i\lt j\le n } 1$$
I found that the inner sum was operated $n$ times then $n-1$ times then $n-2$ times ... so finally it was operated $\sum_{k=1}^{n} k$ = $\frac{(n)(n+1)}{2}$ = $\binom{n+1}{2}$ times, which is the answer.
Using this approach becomes very complicated when higher number of summation signs are involved.
The general solution for $m$ summation signs seems to be $\binom{n+1}{m}$ though I do not know how to show!

REPLY [9 votes]: If we use the formula
$$
\sum_{k=0}^n\binom{k}{m}=\binom{n+1}{m+1}\tag{1}
$$
the method in the question is not all that cumbersome. Each sum building on the previous one, using $(1)$, we start with the sum for $1$ index:
$$
\sum_{k=0}^n\binom{k}{0}=\binom{n+1}{1}\tag{2}
$$
The sum for $2$ indices is
$$
\sum_{k=0}^{n-1}\binom{k+1}{1}=\binom{n+1}{2}\tag{3}
$$
The sum for $3$ indices is
$$
\sum_{k=0}^{n-1}\binom{k+1}{2}=\binom{n+1}{3}\tag{4}
$$
Finally, the sum for $4$ indices is
$$
\sum_{k=0}^{n-1}\binom{k+1}{3}=\binom{n+1}{4}\tag{5}
$$