I remember that problem from university! (But I didn't solve it, somebody gave me a big clue)
The answer is that it is impossible, you can't cover the circle of diameter 20 using only 19 strips $1\times 20$.
To see why, suppose you have such cover and imagine a sphere of diameter 20 cut in half by our circle. Find the orthogonal projection of each strip on the sphere, it is a ring, and it is easy to compute it's area $2\pi R x$, where $R$ is the radius of the sphere and $x$ the width of the strip.
But between all the rings, they cover the whole sphere so the total area must be at least $4\pi R^2$, so
$$ N \times 2 \pi R x \ge 4 \pi R^2 $$
where $N$ is the number of strips. Letting $N=19, R=10$ and $x=1$ we obtain a contradiction.
Notation
I have assumed that “$2\times3$” means “$2$ rows and $3$ columns”. Is my assumption correct?
It depends on the context. In general I'd agree, but since the counts you quote doesn't include a separate handling of the $3\times2$ case, in this situation I'd rather assume that $2\times3$ is meant to cover both orientations, both two rows and three columns and two columns and three rows.
Total count
Can someone help me to find how many rectangles there are in the rectangular grid below?
If all you are interested in is the total number of non-square rectangles, then it doesn't matter how you write down each individual count. In fact, I'd not iterate over all possible shapes, but instead use a different approach.
For a grid of $m\times n$ tiles, there are $(m+1)\times(n+1)$ tile corners. If you consider each of them as a possible corner of a rectangle, and define each rectangle by two opposite corners, then you have a total of
$$\bigl((m+1)(n+1)\bigr)^2$$
possible combinations of two points. Obviously, that number is way too large, because we counted some things we shouldn't, and counted others more than once. So let's address those issues.
You don't want either dimension of such a rectangle to be zero. So if you picked the first point, then there are $m+n+1$ possible positions where that second point would be in the same row and/or the same column as the first. Subtract that and you are at
$$(m+1)(n+1)\bigl((m+1)(n+1)-(m+n+1)\bigr)=(m+1)(n+1)mn\;.$$
As Arthur pointed out in a comment, you can think of this as picking the second point from the $m\times n$ possible corner positions which you obtain by removing the “forbidden” row and column. At this point you have two points in distinct rows and columns, but nothing is sorted yet. So every possible rectangle is counted four times: any of its four corners might be the first point, with the opposite corner as the second point. So the number of rectangles is
$$\tfrac14(m+1)(n+1)mn\;.$$
Now you still have to get rid of those squares. How many are there? Let's look for a pattern. There are $m\times n$ squares of size $1$. There are $(m-1)\times(n-1)$ squares of size $2$, and so on. The maximal size such a square can have is the smaller of the dimensions of your tile grid. So there are
$$\sum_{i=1}^{\min(m,n)}(m+1-i)(n+1-i)$$
squares in total. Assuming $m\le n$ this simplifies to
$$\sum_{i=1}^{m}(m+1-i)(n+1-i)=\tfrac16 m(m+1)(3n-m+1)\;.$$
Now take the number of all rectangles, subtract all squares, and you are done. If you expand and then factor the result, you get a total count of
$$\tfrac1{12}m(m+1)(3n^2 + 2m - 3n - 2)=
\tfrac1{12}m(m+1)\bigl(3n(n-1) + 2(m-1)\bigr)\;.$$
For $m=n=3$ this indeed yields a count of $22$. For $m=5,n=7$ (remember to ensure $m\le n$) this gives a count of $335$ non-square rectangles. You can verify that count with a brute-force computation.
Bonus question
If you want to, it might be interesting to think about why that formula always results in an integer if $m,n$ are integers.
Either $m$ or $m+1$ is divisible by two, so their product is even. For the same reason, $n(n-1)$ is even, and since $2(m-1)$ is even, too, $\bigl(3n(n-1) + 2(m-1)\bigr)$ must be even. So the whole thing is the product of two even numbers, and therefore divisible by four.
Exactly one of $m,(m+1),(m-1)$ is divisible by three. In the first two cases, the whole product is obviously divisible by three. In the last case, $\bigl(3n(n-1) + 2(m-1)\bigr)$ is divisible by three since it's the sum of two integers which are divisible by three. As the whole thing is divisible by four and by three, it is divisible by twelve.
Best Answer
The width must be odd and the height must be odd. There are $10$ choices of a pair of $x$-coordinates to have width $1$, $8$ choices for width $3$, $6$ choices for with $5$, etc. So we have $10+8+6+4+2=30$ ways to choose the two $x$ coordinates and likewise $30$ ways to choose $y$-coordinates. This gives us a total of $30\cdot 30=900$ odd area rectangles.
What you did was to pick one vertex and than assume that each possible odd width and height could be realized with this vertex.