chessboard puzzle

We can view the chessboard as being defined by $9$ evenly spaced vertical lines and $9$ evenly spaced horizontal lines.

Now choose $2$ distinct vertical lines and $2$ distinct horizontal lines. The resulting points of intersection are the vertices of one of the rectangles we are wanting to count. In fact, there is a one-to-one correspondence between these $\binom{9}{2} \binom{9}{2} = 1296$ choices of vertical and horizontal lines and the rectangles in a chessboard, and so we conclude that there are $\boxed{1296}$ rectangles.

There is one more method of solving this question or any other questions where it is a square of n X n sides.

The number of rectangles in n X n square divided into squares of 1 unit is:

$1^{3}$ + $2^{3}$ + $3^{3}$ ....... + $n^{3}$

Since it's a chessboard it will have :

$1^{3}$ + $2^{3}$ + $3^{3}$ + $4^{3}$ + $5^{3}$ + $6^{3}$ + $7^{3}$ + $8^{3}$

= 1+8+27+64+125+216+343+512

= $\boxed{1296}$

Since there are 9 vertical and 9 horizontal lines in a chessboard, then we only need the product of the combinations of the horizontal and vertical lines taken 2 at a time. So, 9C2*9C2 = 1296.

There are 64 one-by-one squares,

49 two-by-two squares, ...

(8-n)^2 n-by-n squares, ...

1 eight-by-eight square;

2 x (7x8) one-by-two rectangles,

2 x (6x8) one-by-three rectangles, ...

2 x (1x8) one-by-eight rectangles;

2 x (6x7) two-by-three rectangles, ...

2 x (1x7) two-by-eight rectangles; ...;

2 x (1x2) seven-by-eight rectangles.

This can all be simplified to find the sum total as the sum of the cubes of integers 1 to 8, which is 8^2 x 9^2 / 4 or 36^2 = 1296.