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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?

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    What does it mean to be different? If they're the same shape but in different places, are they different? – 2017-01-19
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    $2^{100}$ *might* be the answer. – 2017-01-19
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    @O. Von Seckendorff the exact result is that of WiCK3D POiSON – 2017-01-19
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    @JeanMarie I meant $2^{100}$ if we didn't consider rotations. Thanks, however. – 2017-01-19
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    Presumably rectangles that are angled to the checkers are not allowed? – 2017-01-19

4 Answers 4

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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:

$$\frac{^{121}C_2-(2\cdot11\cdot\ ^{11}C_2) }{2}$$

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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 $55^2=3025$.

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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.

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    Shouldn't you add $100$ to account for rectangles that consist of a single square? – 2017-01-19
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    @Barry Yes, I should! – 2017-01-19
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    What if the two points chosen lie on a single edge ? – 2017-01-19
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    Then that is a one-dimensional rectangle. – 2017-01-19
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    Wait, how did you get 2*10-1? – 2017-01-19
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    It's $2*10-2$, the number of squares in a row and a column, without their intersection (you have to subtract the intersection for both the row and the column). – 2017-01-19
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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 $$ \sum_{m=1}^{10}\sum_{n=1}^{10}(11-m)(11-n)=\left(\sum_{m=1}^{10}11-m\right)^2=\left(\sum_{m=1}^{10}m\right)^2=3025. $$

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    You can even solve it using your previous method by excluding all 1-D rectangles and removing the repeated rectangles. You can write it as $$\frac{^{121}C_2-(2\cdot11\cdot\ ^{11}C_2) }{2}$$ – 2017-01-19