# Analysis of how-many-squares and rectangles are are there on a chess board?

I know the formula $$f(n) = \frac{n \cdot (n + 1) \cdot (2n + 1) } 6$$

Since chess board consists $8\times 8$, hence here $n=8$.

But I want to know how it has been concluded? Also how to tackle the number of rectangles?

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Here is an easy way to count the number of rectangles. There are $9$ horizontal lines on the chessboard, and $9$ vertical lines. Choose two distinct horizontal lines, and two distinct vertical lines. These determine a unique rectangle. And any rectangle determines a pair of horizontal lines and a pair of vertical lines.

So the number of rectangles is $\binom{9}{2}^2$. That is $1296$.

Exactly the same idea can be used to count the number of rectangles on an $m$ by $n$ chessboard. It is $$\binom{m+1}{2}\binom{n+1}{2}.$$

The number of squares is a bit less nice. It is easy to see that there are $8^2$ small $1\times 1$ squares, $7^2$ $2\times 2$ squares, and so on down to $1^2$ $1\times 1$ squares, for a total of $$1^2+2^2+3^2+\cdots+8^2.$$ Now we can add up, but there is also a simple formula for the sum of the first $n$ squares. The same idea works for counting the squares on an $n \times n$ chessboard.

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It is easy to see that there will be 64 small (1 x1) squares . But what is the logic behind 2 x 2 squares ? How is it $7^2$ – Geek Aug 4 '12 at 7:36
@Geek, consider for $2 \times 2$ squares how many centers they have. – ladaghini Aug 4 '12 at 7:45
@ladaghini not an elegant way to think in terms of center ...How would you calculate for 3 x 3 then ? – Geek Aug 4 '12 at 7:47
@Geek: For the $3\times 3$, for example, consider the top left corner. It can take $6$ positions on the top line, $6$ positions on the second line, and so on up to $6$ positions on the sixth line, total $6^2$. Equivalently, the top left little square of the $3\times 3$ can take $6$ positions on the first row, $6$ positions on the second row, and so on. Same idea works for $k\times k$ squares. – André Nicolas Aug 4 '12 at 7:52
@mehdi: consider this picture. – robjohn Jun 16 '14 at 15:25

If you look at a chessboard ($n = 8$), there are

• 1 square of size 8x8
• 4 squares of size 7x7
• 9 squares of size 6x6
• and so on until
• 64 squares of size 1x1

So, really, the formula boils down the the summation $\sum_{i=1}^n i^2$, which can be derived as shown in the answers here. See also a combinatorial approach to better suit your question.

For rectangles, note that for an $n \times m$ chessboard, there are $n+1$ lines that form the rows and $m+1$ lines that form the columns. A rectangle will be bound by two of these rows and two of these columns, so the number of rectangles is given by:

$R(n,m) = \binom{n+1}{2}\binom{m+1}{2}$

Note that the formula is generalized for any rectangular chessboard, not just a square one.

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To get the squares, integrate $(m+1-x)(n+1-x)$ for $x$ from $1$ to $m$, where $m$ and $n$ are the dimensions of a $m\times n$ chessboard, where $m\lt n$.

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