There is a rectangular sheet of dimension (2m | vedclass

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(D) rectangle is formed by choosing two horizontal lines and two vertical lines. Let the horizontal lines be $y_0, y_1, \dots, y_{2n-1}$ and vertical

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$m^2n^2$

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(D) rectangle is formed by choosing two horizontal lines and two vertical lines. Let the horizontal lines be $y_0, y_1, \dots, y_{2n-1}$ and vertical lines be $x_0, x_1, \dots, x_{2m-1}$. $A$ rectangle has odd side lengths if the difference between the indices of the chosen lines is odd. For the horizontal side,we need to choose two lines $y_i, y_j$ such that $|i - j|$ is odd. This means one index must be even and the other must be odd. In the set ${0, 1, \dots, 2n-1}$,there are $n$ even numbers ${0, 2, \dots, 2n-2}$ and $n$ odd numbers ${1, 3, \dots, 2n-1}$. The number of ways to choose one even and one odd index is $n 	imes n = n^2$. Similarly,for the vertical side,the number of ways to choose two lines such that the difference is odd is $m 	imes m = m^2$. Therefore,the total number of rectangles with odd side lengths is $m^2 	imes n^2 = m^2n^2$.

There is a rectangular sheet of dimension $(2m - 1) \times (2n - 1)$,(where $m > 0, n > 0$). It has been divided into squares of unit area by drawing lines perpendicular to the sides. Find the number of rectangles having sides of odd unit length.

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