## In a $3 \times 2$ cross-hatched grid, a total of $37$ different rectangles could be situated within that grid as indicated in the sketch.

There are $5$ grids smaller than $3 \times 2$, vertical and horizontal dimensions being important, i.e. $1 \times 1$, $2 \times 1$, $3 \times 1$, $1 \times 2$ and $2 \times 2$. If each of them is cross-hatched, the following number of different rectangles could be situated within those smaller grids:

$1 \times 1$$1$

$2 \times 1$$4$

$3 \times 1$$8$

$1 \times 2$$4$

$2 \times 2$$18$

Adding those to the $37$ of the $3 \times 2$ grid, a total of $72$ different rectangles could be situated within $3 \times 2$ and smaller grids.

How many different rectangles could be situated within $47 \times 43$ and smaller grids?

### Let’s consider a grid of size $m \times n$.

The total number of rectangles in such a grid would be the sum of all possible rectangles of size $1 \times 1$, $2 \times 1$, $2 \times 2$, …, up to size $m \times n$.

To count all possible rectangles of size $p \times q$ within an $m \times n$ grid, we can choose $p$ rows out of $m$ and $q$ columns out of $n$. The number of ways we can choose $p$ items from $m$ items is given by the combination formula $mCp = m! / (p!(m-p)!)$. Similarly, the number of ways we can choose $q$ items from $n$ items is $nCq = n! / (q!(n-q)!)$.

Therefore, the total number of rectangles of size $p \times q$ within a $m \times n$ grid is $mCp * nCq$.

The total number of rectangles within an $m \times n$ grid would be the sum of rectangles of all possible sizes. We need to sum over all $p \le m$ and $q \le n$.

$$\textrm{Total rectangles} = \sum_{p=1}^{m} \sum_{q=1}^{n} (mCp * nCq)$$

To calculate the total number of rectangles in a $47 \times 43$ grid and smaller, we need to sum over all possible grid sizes from $1 \times 1$ to $47 \times 43$.

Notice that the problem becomes symmetric if we sum over $m$ and $n$ separately.

Thus, the total number of rectangles in $47 \times 43$ and smaller grids would be:

$$\textrm{Total rectangles} = \sum_{m=1}^{47} \sum_{n=1}^{43} \sum_{p=1}^{m} \sum_{q=1}^{n} (mCp * nCq)$$

Please note that this is a complex summation to compute and will take some time. It may be best done by a computer program.

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