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

Options

A. `(m+n+1)^(2)`
B. `mn(m+1)(n+1)`
C. `m^(m+n-2)`
D. `m^(2)n^(2)`

Explanation

To solve the problem of finding the number of rectangles with odd unit lengths in a rectangular sheet of dimensions \((2m-1) \times (2n-1)\), we can follow these steps:

### Step 1: Understand the dimensions of the rectangle
The rectangle has dimensions \(2m-1\) and \(2n-1\). This means that the rectangle can be divided into squares of unit area, resulting in a grid of \(2m-1\) rows and \(2n-1\) columns.

### Step 2: Count the number of horizontal and vertical lines
Since the rectangle is divided into unit squares, the number of horizontal lines is \(2m\) (one line at each unit length from 0 to \(2m-1\)) and the number of vertical lines is \(2n\) (one line at each unit length from 0 to \(2n-1\)).

### Step 3: Identify odd-length rectangles
To form a rectangle with odd-length sides, we need to select lines that correspond to odd indices. The odd indices in the range of \(2m\) horizontal lines are \(1, 3, 5, \ldots, (2m-1)\), and similarly for the \(2n\) vertical lines.

### Step 4: Count the odd lines
The number of odd horizontal lines from \(0\) to \(2m\) is \(m\) (since the odd numbers up to \(2m\) are \(1, 3, 5, \ldots, (2m-1)\)). The same applies for the vertical lines, so there are \(n\) odd vertical lines.

### Step 5: Calculate the number of ways to choose lines
To form a rectangle, we need to choose 2 horizontal lines and 2 vertical lines. The number of ways to choose 2 lines from \(m\) odd horizontal lines is given by the combination formula \(C(m, 2)\), and similarly for the \(n\) vertical lines, it is \(C(n, 2)\).

The formula for combinations is given by:
\[ C(k, 2) = \frac{k(k-1)}{2} \]

### Step 6: Calculate the total number of rectangles
Thus, the total number of rectangles with odd-length sides can be calculated as:
\[ \text{Total Rectangles} = C(m, 2) \times C(n, 2) = \left(\frac{m(m-1)}{2}\right) \times \left(\frac{n(n-1)}{2}\right) \]

### Step 7: Simplify the expression
This simplifies to:
\[ \text{Total Rectangles} = \frac{m(m-1) \cdot n(n-1)}{4} \]

### Final Answer
Hence, the number of rectangles having sides of odd unit length is:
\[ \frac{m(m-1) \cdot n(n-1)}{4} \]