About checkered rectangulars There is a  10 X 10 checkered square. How many different checkered rectangulars can one find on the square?
Must I find the amount of ways for 1*10, 2* 10...etc and summarize them or is there some other way?
 A: Hint:
A unique rectangle can be formed by choosing any 2 sides from 11 length sides and 2 sides from 11 breadth sides.

 Solution: 
$$^{11}C_2\cdot ^{11}C_2$$

Edit (Alternate method):
As MichaelBurr suggested that a rectangle can be defined by its LLC(Lower Left Corner) and URC(Upper Right Corner) or LRC and ULC, we can count the number of rectangles by choosing any $2$ points from $121$ points(i.e $^{121}C_2$).
But in the above pairs of points chosen, there are points which may lie on same edge, excluding them i.e. $2\cdot 11\cdot\ ^{11}C_2$, since there are $22$ edges that make up the $10\times 10$ checkerboard, and each edge has $11$ points on it. 
The remaining rectangles left are counted twice because the points' pair can be LLC and URC or LRC and ULC. Hence total number of triangles is:
$$\frac{^{121}C_2-(2\cdot11\cdot\  ^{11}C_2) }{2}$$
A: Any rectangle is individuated by the coordinates of its diagonal corners
$$
\left( {x_{\,1} ,y_{\,1} } \right)\left( {x_{\,2} ,y_{\,2} } \right)
$$
with
$$
\left\{ \begin{gathered}
  0 \leqslant x_{\,1}  < x_{\,2}  \leqslant 10 \hfill \\
  0 \leqslant y_{\,1}  < y_{\,2}  \leqslant 10 \hfill \\ 
\end{gathered}  \right.
$$
The $x$es can be choosen in 
$$
1 \cdot 10 + 1 \cdot 9 +  \cdots 1 \cdot 1 = \frac{{10 \cdot 11}}
{2} = 55
$$
ways
and same for the $y$s, which can be choosen indipendently.
So the answer is $55^2=3025$.
A: Hint: If location matters (the same shape in two different places counts as different rectangles), then a rectangle be determined by its two corners.  We need to specify either (the upper left and lower right corner) or (the upper right and lower left corner).
There are a few special cases that must be considered:


*

*For a singleton, all of the corners are the same.  There are $10^2=100$ such rectangles.

*For a rectangle where one dimension is $1$, then the two pairs above are the same.  For this case, there are $10^2=100$ choices for the first corner and then $2\cdot 10-2=18$ choices for the second corner.  This gives $\frac{1}{2}\cdot 100\cdot18=900$ possibilities (the $\frac{1}{2}$ accounts for reversing the order of the choice of corners).

*For a rectangle where neither of the dimensions is $1$, then we can choose the two corners as $100$ choices for the first corner and $100-20+1=81$ choices for the second corner.  This results in $\frac{1}{4}\cdot 100\cdot 81=2025$ rectangles.  We must divide by $4$ because we could choose the corners in either order or choose the alternate pair of corners.
Adding these up, I get $2025+900+100=3025$ different rectangles.
A: There are $10$ possible side lengths for each dimension of a rectangle.  Since the upper left corner of a rectangle of dimensions $m\times n$ must be at least $m-1$ rows from the bottom and $n-1$ columns from the right edge, this means that there are $(11-m)\times(11-n)$ possible positions for the upper left corner.  This means that the number of rectangles is
$$
\sum_{m=1}^{10}\sum_{n=1}^{10}(11-m)(11-n)=\left(\sum_{m=1}^{10}11-m\right)^2=\left(\sum_{m=1}^{10}m\right)^2=3025.
$$
