Given a $n{\times}n$ matrix $A$. $A$ can be partitioned into cells in $2^{2n-2}$ ways.


A $n{\times}n$ matrix has $n^2$ elements. The number of columns is $n$, and the number of rows is $n$. A partitioning of the matrix into cells is done by drawing lines to divide the matrix.
The number of horizontal lines is $n-1$.
The number of vertical lines is also $n-1$.
The total number of lines is $2(n-1) = 2n -2$ lines.

There is a partition that involves none of these lines, which is $A$ itself.

All the other partitions are formed by selecting any $k$ lines.

Number of partitions $N$: $$N = \sum_{i=0}^{2n-2} {(2n-2)\choose{i}} \tag{1}$$
Now, the cardinality of a power set $P(S)$ of $S$, where ${\#}S = m$ is given by the below formula:
$${\#}P(S) = 2^m \tag{2}$$
${\#}P(S)$ can alternatively be calculated as:
$${\#}P(S) = \sum_{i=0}^{m} {m\choose{i}} \tag{3}$$ Equivocating $(2)$ and $(3)$: $$\sum_{i=0}^{m} {m\choose{i}} = 2^m \tag{4}$$ Substituting $(4)$ into $(1)$: $$N = 2^{2n -2}$$.


Yes, I think you're correct. Alternatively, once you have established that there are $2n-2$ possible lines, you could argue that each line can either be drawn or not ($2$ choices), and each set of choices of which lines to draw results in a unique partitioning. Thus, there are $2^{2n-2}$ partitions.


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