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How many rectangles are there on an $8 \times 8$ checkerboard?

\begin{array}{|r|r|r|r|r|r|r|r|} \hline & & & & & & & \\ \hline & & & & & & & \\ \hline & & & & & & & \\ \hline & & & & & & & \\ \hline & & & & & & & \\ \hline & & & & & & & \\ \hline & & & & & & & \\ \hline & & & & & & & \\ \hline \end{array}

Attempt

I just counted them up via casework: a $1 \times 1: 64 $

$1 \times 2: 56 $

$\vdots$

$1 \times 8: 8$ Then,

$2 \times 2: 49$

$2 \times 3: 42 $

$\vdots$

The pattern continues like it looks like it should.

Thus, we can sum up all of these solutions as $8(1+\cdots+8)+7(1+\cdots+7)+\cdots+2(1+2)+1(1) = 750$, but the correct answer is $1296$. Where did I go wrong?

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  • $\begingroup$ Just as a comment, it's a little weird to me that the question which is interesting and well written has 3 upvotes while the answer (again well done) has 10 upvotes. Who are the people that upvoted the answer and not the question? Please consider giving the question some love too. $\endgroup$ – user223391 Dec 23 '15 at 3:54
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First, the easy way to count them is to notice that each rectangle is completely determined by its top and bottom edges and its left and right edges. Pick any two of the nine horizontal lines and any two of the nine vertical lines, and you’ve picked out a rectangle. Conversely, each rectangle determines two horizontal and two vertical lines, those on which its edges lie. Thus, there must be

$$\binom92^2=36^2=1296$$

rectangles.

Your calculation is off because you forgot that a rectangle can be wider than it is tall, so you missed half of the non-square rectangles.

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