How many rectangles, whose boundary equals or exceeds that of a $3{\times}3$ square, can fit into a $9{\times}9$ square. I want to know how many rectangles can fit inside a $9$ by $9$ square.
However, I want to exclude any rectangles that are not greater than or equal to a $3$ by $3$ square. 
By this, I mean excluding rectangles such as $(2*3),(3*2),(3*1),(3*1),(2*2)$...and so on because the are not equal to, or can be fully surrounded by, a $3*3$ square.
I have arrived at a figure of $784$ by visualizing the problem and crunching the numbers, but I would like a sounder theoretical approach that makes sense.
 A: This answer assumes:


*

*Two different e.g. $5 \times 6$ rectangles at different positions (within the $9 \times 9$ grid) are counted as distinct.

*You want to exclude e.g. $7 \times 2$ rectangles since such a rectangle is "not greater than" (in the sense it cannot enclose) a $3 \times 3$ square.  In other words, you want to count $m \times n$ rectangles where $m \ge 3$ and $n \ge 3$.
Here is a way to count the above without exhaustive listing. 
The $9\times 9$ grid of $81$ squares has $10$ vertical lines and $10$ horizontal lines, and a rectangle is defined by a pair of vertical lines and a pair of horizontal lines.  
The number of ways to choose the pair of horizontal lines is ${10 \choose 2} = 45$ if you ignore the size constraint.  However, if you require $m \ge 3$, then you must exclude pairs which are one square apart ($9$ such pairs) or two squares apart ($8$ such pairs).  So the number of pairs satisfying $m \ge 3$ is $45 - 9 - 8 = 28$
Similarly the number of pairs of vertical lines satisfying $n \ge 3$ is also $28$.
Thus the number of valid rectangles is $28^2 = 784$.
A: The way I did it before reading Antkam's much better solution was:
A 3*3 cell in the top right has 1 position + 6 positions to the right. (7 total) 
A 3*4 cell in the top right has 1 position + 5 positions to the right. (6 total) 
A 3*5 cell in the top right has 1 position + 4 positions to the right. (5 total) 
A 3*6 cell in the top right has 1 position + 3 positions to the right. (4 total) 
A 3*7 cell in the top right has 1 position + 2 positions to the right. (3 total) 
A 3*8 cell in the top right has 1 position + 1 positions to the right. (2 total)
A 3*9 cell in the top right has 1 position + 0 positions to the right. (1 total) 
This means we have (7+6+5+4+3+2+1) = 28 rectangles in the top three rows alone.
Now repeat the process on the first three columns for another 28 rectangles.
Then there must be 28*28 total rectangles formed by the intersection of these two groups of rectangles along the top and left hand side.
