Question

The number of rectangles which are not squares having common side in a chess board is
(a) ${}^{8}{{C}_{2}}\times {}^{8}{{C}_{2}}-{{8}^{2}}$
(b) $\left( {}^{8}{{C}_{2}}\times {}^{8}{{C}_{2}} \right)-\sum{{{8}^{2}}}$
(c) ${}^{9}{{C}_{2}}\times {}^{9}{{C}_{2}}-{{8}^{2}}$
(d) $\left( {}^{9}{{C}_{2}}\times {}^{9}{{C}_{2}} \right)-\sum{{{8}^{2}}}$

Answer 1

We will first calculate the number of all possible rectangles on the chess board. Then we will find the number of all possible squares on the chess board. Then the number of rectangles that are not squares will be the difference between those two.

Complete step-by-step answer:
There are 9 vertical lines and 9 horizontal lines on a chess board. We can form a rectangle on the chess board by choosing any 2 vertical lines and any two horizontal lines. So, the total number of rectangles possible on the chess board is ${}^{9}{{C}_{2}}\times {}^{9}{{C}_{2}}$.
Now, we will calculate the number of all possible squares on the chess board. If we fix the distance between the vertical lines and horizontal lines while choosing to be 1, we will get all the single squares on the chess board. The single squares are $64={{8}^{2}}$. Now, for squares having 4 single units, or width 2, we will fix the distance between the vertical lines and horizontal lines while choosing to be 2. So, we will get ${{7}^{2}}$ such squares. Continuing this pattern, the total number of squares will be
${{8}^{2}}+{{7}^{2}}+{{6}^{2}}+\cdots {{1}^{2}}$ . Let us denote this sum by $\sum{{{8}^{2}}}$.
Hence, the number of rectangles that are not squares is ${}^{9}{{C}_{2}}\times {}^{9}{{C}_{2}}-\sum{{{8}^{2}}}$.

So, the correct answer is “Option (d)”.

Note: Choosing the setup correctly to use the combinations formula is very crucial. Every case needs to be taken into account. Missing a single trivial case can lead to an incorrect calculation. To answer questions with a similar type of approach for the solution needs practice in handling the combinations formula.