Question

Number of rectangles in the grid has shown which are not squares?

(a) 160
(b) 162
(c) 170
(d) 185

Answer 1

Hint: A rectangle is constructed of 4 vertices so we are going to select two vertices from the 7 vertices which are lying horizontally and selecting 2 vertices from 5 vertices which are lying vertically and then multiply both of the selections. Now, we got the number of rectangles possible but in these rectangles, we got squares also so subtracting squares from these rectangles. Number of squares that could be possible in the above grid is calculated using the formula for $n\times m$ grid is equal to $n\left( m \right)+\left( n-1 \right)\left( m-1 \right)+\left( n-2 \right)\left( m-2 \right)+...\left( n-n \right)\left( m-n \right)$ . Now, put n as 4 and m as 6 in this formula to get the number of squares.

Complete step-by-step solution:
The grid in the above problem is given as:

As you can see that in the above grid we have 4 rows and 6 columns so the grid is of $4\times 6$ order. Also, there are 7 horizontal vertices and 5 vertical vertices.
We are asked to find the number of rectangles in this grid, which are not squares. First of all, we are going to find all the rectangles that could be constructed in this grid which are squares also. For that, we are selecting two vertices from 7 horizontal vertices and selecting 2 vertices from 5 vertical vertices by combinatorial method then we will multiply these selections.
${}^{7}{{C}_{2}}\left( {}^{5}{{C}_{2}} \right)$
We know that,
${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
So, using this relation to simplify the above combinatorial expression we get,
$\begin{align}
  & \dfrac{7!}{2!\left( 7-2 \right)!}\left( \dfrac{5!}{2!\left( 5-2 \right)!} \right) \\
 & =\dfrac{7!}{2!5!}\left( \dfrac{5!}{2!3!} \right) \\
 & =\dfrac{7.6.5!}{2!5!}\left( \dfrac{5.4.3!}{2!3!} \right) \\
\end{align}$
In the above expression, $5!\And 3!$ will be cancelled out from the numerator and denominator and we get,
$\begin{align}
  & \dfrac{7.6}{2.1}\left( \dfrac{5.4}{2.1} \right) \\
 & =\dfrac{42}{2}\left( \dfrac{20}{2} \right) \\
 & =210 \\
\end{align}$
Now, we are going to find the number of squares that could be possible in this $4\times 6$ order grid which we are going to find by using the formula for $n\times m$ grid which is equal to $n\left( m \right)+\left( n-1 \right)\left( m-1 \right)+\left( n-2 \right)\left( m-2 \right)+...\left( n-n \right)\left( m-n \right)$.
Substituting n as 4 and m as 6 in the above formula we get,
$\begin{align}
  & 4\left( 6 \right)+\left( 4-1 \right)\left( 6-1 \right)+\left( 4-2 \right)\left( 6-2 \right)+\left( 4-3 \right)\left( 6-3 \right)+\left( 4-4 \right)\left( 6-4 \right) \\
 & =24+3\left( 5 \right)+2\left( 4 \right)+\left( 1 \right)\left( 3 \right)+0\left( 2 \right) \\
 & =24+15+8+3+0 \\
 & =50 \\
\end{align}$
Hence, we got the number of squares as 50. Now, subtracting 50 from 210 we will get the number of rectangles which are not squares.
$\begin{align}
  & 210-50 \\
 & =160 \\
\end{align}$
Hence, the correct option is (a).

Note: While reading the question it might come to your mind first to count the number of rectangles of the given grid. This is not the wrong way of approaching the problem but in this way, there is a high probability that you might miss out on some of the rectangles which are not squares so it will be more beneficial to use the way that we have shown above.