Do direct products commute with the direct sums of vector spaces?
Basically is $\underset{i \in I}{\prod} \underset{j \in J}{\bigoplus}M_{i,j} \cong \underset{j \in J}{\bigoplus}\underset{i \in I}{\prod} M_{i,j}$?
What if $M_{i,j} \cong M_{\tilde{i},\tilde{j}}$ for each $i,\tilde{i}\in I$ and $j, \tilde{j}\in J$?
If you ask for a random isomorphism, this is a question of cardinalities, and therefore rather boring. It is more natural to ask if the canonical homomorphism (it exists by the universal properties)
$\oplus_j \prod_i M_{i,j} \to \prod_i \oplus_j M_{i,j}$
is an isomorphism. Clearly this is the case when $I$ or $J$ are finite. In general, even for modules $M_{i,j}$, it is clearly a monomorphism (both modules embed into $\prod_i \prod_j M_{i,j}$), but not an epimorphism when $I,J$ are infinite and $M_{i,j} \neq 0$ for all $i,j$:
Choose elements $0 \neq m_{i,j} \in M_{i,j}$. Choose injections $u : \mathbb{N} \to I$ and $v : \mathbb{N} \to J$. For every $n \in \mathbb{N}$ let $x_{u(n)} = m_{u(n),v(n)} \in \oplus_j M_{u(n),j}$. For $i \notin \mathrm{im}(u)$ let $x_i=0$. Then $x=(x_i)_i \in \prod_i \oplus_j M_{i,j}$. It has no preimage in $\oplus_j \prod_i M_{i,j}$, since by construction infinitely many $j$s are used.