Question

A rectangle with sides of length $ (2m – 1)$ and $ (2n – 1)$ units is divided into squares of unit length by drawing parallel lines as shown in the diagram, then the number of rectangles possible with odd side lengths is equal to:

• A. ${(m+n-1)}^2$
• B. $4^{m+n-1}$
• C. $m^2{n}^2$
• D. $m(m+2)n(n+1)$

Answer 1

The correct option is C

$m^2{n}^2$

Explanation for the correct answer:

Using permutation to find the number of rectangles:

Total number of vertical lines $=2n$

Total number of horizontal lines $=2m$

For the number of rectangles possible with odd side lengths, we will take one odd side and one even so that their difference will give us an odd number

Therefore, the number of rectangles is

$$\begin{gathered} ={}^{m}C_1.{}^{m}C_1.{}^{n}C_1.{}^{n}C_1 \\ =m^2{n}^2\left[\because {}^n{C}_1=n\right] \end{gathered}$$

Therefore, the correct answer is option (C).