Question

The number of rectangles excluding squares from a rectangle of size $9 \times 6$ is
A. 391
B. 791
C. 842
D. 250

Answer 1

Hint:
We will have 10 vertical lines and 7 horizontal lines for a rectangle of grid size $9 \times 6$. We will first count the number of rectangles possible using combination. Then, we will subtract the number of possible squares.

Complete step by step solution:
If we are given that the grid size of the rectangle is $9 \times 6$.
For a rectangle of grid size $9 \times 6$, we will require 10 vertical lines and 7 horizontal lines.

For one rectangle, we have to choose 2 vertical lines and 2 horizontal lines.
Hence, the total number of rectangles are
${ = ^{10}}{C_2}{ \times ^7}{C_2}$
And we know that $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
On simplifying it further, we will get,
$
   = \dfrac{{10!}}{{2!8!}} \times \dfrac{{7!}}{{2!5!}} \\
   = \dfrac{{10.9.8!}}{{2.1.8!}} \times \dfrac{{7.6.5!}}{{2.1.5!}} \\
   = 45 \times 21 \\
   = 945 \\
$
But, we have to exclude squares from it.
Taking the size of the square box of unit size, then we have $9 \times 6$ boxes.
If the size of the square box is 2 units, then the number of boxes is $8 \times 5$
If the size of the square box is 3 units, then the number of boxes is $7 \times 4$
If the size of the square box is 4 units, then the number of boxes is $6 \times 3$
If the size of the square box is 5 units, then the number of boxes is $5 \times 2$
If the size of the square box is 6 units, then the number of boxes is $6 \times 1$
Then, the total number of squares is:
 $
   = \left( {9 \times 6} \right) + \left( {8 \times 5} \right) + \left( {7 \times 4} \right) + \left( {6 \times 3} \right) + \left( {5 \times 2} \right) + \left( {4 \times 1} \right) \\
   = 54 + 40 + 28 + 18 + 10 + 4 \\
   = 154 \\
$
 Then, the number of rectangles excluding squares is $945 - 154 = 791$

Therefore, option B is correct.

Note:
When we count the numbers of squares for $m$ lines and $n$ lines, then there will be $n - 1$ type of unit boxes if $m > n$. Similarly, if $n > m$, then there will be $m - 1$ types of unit squares.