Finite Summations can be Interchanged In the proof of associativity of matrix multiplication, the reason for one of the steps is given as -Finite summations can be interchanged. What is meant by this statement?

 A: In other words, they mean the following: if $c_{ij}$ is a number for every $i = 1,\dots,m$ and $j = 1,\dots,n$, then
$$
\sum_{i=1}^m \left(\sum_{j=1}^n c_{ij}\right) =  \sum_{j=1}^n \left(\sum_{i=1}^m c_{ij}\right).
$$
Note that in most texts, the parentheses around the inner sum are not written explicitly (but are implied).
That these sums are equal is an "obvious" consequence of the commutativity of addition.
For instance, we can compare the two expressions in the case of $m = 2$ and $n = 3$. We have
$$
\sum_{i=1}^2 \sum_{j=1}^3 c_{ij} = \sum_{i=1}^2 (c_{i1} + c_{i2} + c_{i3}) = (c_{11} + c_{12} + c_{13}) + (c_{21} + c_{22} + c_{23}).
$$
On the other hand ,
$$
\sum_{j=1}^3 \sum_{i=1}^2 c_{ij} = \sum_{j=1}^3 (c_{1j} + c_{2j}) = 
(c_{11} + c_{21}) + (c_{12} + c_{22}) + (c_{13} + c_{23}).
$$
Clearly, since the same terms are being added in both sums, the two sums are equal.

The particular step, in detail:
$$
\begin{align}
\sum_{j=1}^n a_{ij}\left(\sum_{k=1}^p b_{jk}c_{kl}\right) &= 
\sum_{j=1}^n \left(\sum_{k=1}^p a_{ij}b_{jk}c_{kl}\right) 
\\ &= 
\sum_{k=1}^p \left(\sum_{j=1}^n a_{ij}b_{jk}c_{kl}\right) 
= 
\sum_{k=1}^p \left(\sum_{j=1}^n a_{ij}b_{jk}\right)c_{kl}.
\end{align}
$$
A: They probably mean that a formula such as
$$\sum_{k=k_0}^{k_1}\sum_{j=j_0(k)}^{j_1(k)} t_{jk}$$ can usually be rewritten in a form
$$\sum_{j=j_0}^{j_1}\sum_{k=k_0(j)}^{k_1(j)} t_{jk}$$ (the summation symbols have been swapped.)
The simplest case is$$\sum_{k=1}^m\sum_{j=1}^{n} t_{jk}=\sum_{j=1}^{n}\sum_{k=1}^m t_{jk}.$$
This is true because addition is commutative.
