Solving equations of nested sums. I'm new to summation signs and I've been dealing a lot with questions containing nested summation signs. I understand how summation works using sigma signs, I just have no clue how to solve this equation to give you a number.
$$
\sum_{m_1=0}^{9}\sum_{m_2=0}^{m_1-1}\sum_{m_3=0}^{m_2-1}\sum_{m_4=0}^{m_3-1}m_4=x
$$
I know the answer to this question is $x=252$ from Desmos (It's the only calculator I have that does summation like this) but I don't know what steps have to be taken to give $x=252$.
 A: 
We can simplify this multiple sum by writing $m_4$ as sum: $m_4=\sum_{m_5=0}^{m_4-1}1$. We obtain
\begin{align*}
\color{blue}{\sum_{m_1=0}^9}\color{blue}{\sum_{m_2=0}^{m_1-1}\sum_{m_{3}=0}^{m_{2}-1}\sum_{m_{4}=0}^{m_{3}-1}m_{4}}
&=\sum_{m_1=0}^{9}\sum_{m_2=0}^{m_1-1}\sum_{m_{3}=0}^{m_{2}-1}\sum_{m_{4}=0}^{m_{3}-1}\sum_{m_{5}=0}^{m_{4}-1}1\\
&=\sum_{0\leq m_5<m_4<m_3<m_2<m_1\leq 9}1\tag{1}\\
&=\binom{10}{5}\tag{2}\\
&\,\,\color{blue}{=252}
\end{align*}

Comment:

*

*In (1) we use another typical index notation, namely writing the range of summation as inequality chain.


*In (2) we observe the index range contains all ordered $5$-tuples from $\{0,1,2,\ldots,8,9\}$. The number of ordered $5$-tuples is given by the binomial coefficient $\binom{10}{5}$.
A: In general, I would start from the innermost summation, i.e, $\sum_{m_4=0}^{m_3-1}m_4$, do its summation to get a result in terms of $m_3$, then use its result for the next outermost summation (i.e., determine a sum in terms of $m_2$), etc., until you have done all of the summations. However, you also need to be careful with the limits. For example, the innermost one goes from $m_4 = 0$ to $m_4 = m_3 - 1$, but the next outer one starts at $m_3 = 0$ with which the innermost lower bound is $m_4 = 0$ but its upper bound is $m_4 = 0 - 1 = -1 \lt 0$, so there is actually no summation for $m_3 = 0$ in that next outer summation. You therefore need to ensure in your evaluations that you account for this. Note this issue also applies to the next $2$ summations.
A: You can profitably use these conversion
$$
\eqalign{
  & \sum\limits_{m_{\,4}  = 0}^{m_{\,3}  - 1} {m_{\,4} }  = \sum\limits_{m_{\,4}  = 0}^{m_{\,3}  - 1} {\left( \matrix{
  m_{\,4}  \cr 
  1 \cr}  \right)}  = \sum\limits_{\left( {0\, \le } \right)\,m_{\,4} \,\left( { \le \,m_{\,3}  - 1} \right)\,} {\left( \matrix{
  m_{\,3}  - 1 - m_{\,4}  \cr 
  m_{\,3}  - 1 - m_{\,4}  \cr}  \right)\left( \matrix{
  m_{\,4}  \cr 
  m_{\,4}  - 1 \cr}  \right)}  =   \cr 
  &  = \left( \matrix{
  m_{\,3}  \cr 
  m_{\,3}  - 2 \cr}  \right) \cr} 
$$
where:
 - in the 2nd step we have transformed the sum bounds to become implicit in the first binomial;
 - in the 3rd step we have used the "double convolution " formula for binomials.
You can then continue to do the same for the outer sums.
