Chess Board: How to find Number of Squares and Rectangles

This concept is important for CAT and MBA entrance exams.

We all have encountered a frequently asked question, "How many squares are there in a 8*8 chessboard?"

We usually think the answer is 8*8=64, right?

But in this case we have just counted 1*1 squares. What about the 2*2 squares, 3*3 squares, 4*4 squares and so on?

Number of 1*1 squares= 8*8=64

Number of 2*2 squares= 7*7=49

Number of 3*3 squares= 6*6=36

Number of 4*4 squares= 5*5=25

Number of 5*5 squares= 4*4=16

Number of 6*6 squares= 3*3=9

Number of 7*7 squares= 2*2=4

Number of 8*8 squares= 1*1=1

Total number of Squares= 82+72+62+...+22+12= 204

Can you see a pattern?

In a 8*8 chessboard, the total number of squares is ∑82

We can generalize this in the following way:

Total number of squares in a n*n chessboard will be = ∑n2; n varying from 1 to n.

Now let us calculate the number of rectangles in 8*8 chessboard.

A rectangle can have the following dimensions: 1*1, 1*2, 1*3, 1*4… 1*8, 2*2, 2*3, 2*4, …2*8, 3*3, 3*4, ….7*8, 8*8.

We can approach the problem in the following way:

To form a rectangle, we need 4 lines. (2 sets of parallel lines). We can select these lines by = 9C2 * 9C2 = 36*36=1296.

We can generalize this in the following way:

Total number of rectangles in a n*n chessboard will be
= n+1C2*n+1C2

Note: The squares will also be included in counting rectangles.

CAT 19 Demo Lecture

Calculating it every time is cumbersome. Also, we can be asked to calculate the number of squares and rectangles in a m*n board. In this case, the above-derived formulas won't work.

Having a standard formula to calculate the number of squares and rectangles in an m*n chessboard will simplify the problem.

Let us see how we can get a generalized formula to calculate the number of squares and rectangles in an m*n chessboard.

In a m*n board,

Total number of rectangles in a m*n board
= m+1C2*n+1C2

(A rectangle can be formed by selecting 2 lines from m+1 lines and 2 lines from n+1 lines) For example, number of rectangles in a 2*3 board will be =18

Total number of squares in a m*n board= ∑ (m*n); m, n varying from 1 to m,n respectively.

For example, number of squares in 2*3 board = 2*3+1*2=8

For your practice, you can calculate the number of squares and rectangles in a 6*7 board.

Let's solve some examples based on this concept.

Suggested Reading :
Learn the Concept of Painted Sides of a Cube Agreement
Solved Examples

Example 1: In how many ways can you place 2 rooks on 8*8 chessboard such that they are not in attacking positions?

Solution:

Number of ways of selecting 1st rook= 64C1

Number of ways of selecting 2nd rook (it should not be in the same row or column) =7*7/2

Therefore, total number of ways= 64*7*7/2= 32*49=1568.

Example 2: If two squares are chosen on a 8*8 chessboard, what is the probability that they have one side in common?

Solution:

Total number of ways of selecting 2 squares= 64C2=2016

To count the favourable cases, we will consider 3 cases.

Case 1: The corner squares.

There are 4 corner squares. For each corner square, we can select the other square in 2 ways.

Therefore, for this cases, favourable cases= 4*2=8

Case 2: The squares on the edges.

There are 24 squares on the edges. For each such square, we can select the other square in 3 ways.

Therefore, for this cases, favourable cases= 24*3=72

Case 3: The inner squares.

There are 36 inner squares. For each such square, we can select the other square in 4 ways.

Therefore, for this case, favourable cases= 36*4=144

On adding these three cases we get 8+72+144=224

But this is not the final answer.

We have counted every case twice.

So total favourable cases= 224/2=112

The probability= 112/2016

Example 3: If two squares are chosen on a 8*8 chessboard, what is the probability that they have one vertex in common?

Solution:

Total number of ways of selecting 2 squares= 64C2= 2016

To count the favourable cases, we will consider 3 cases.

Case 1: The corner squares.

There are 4 corner squares. For each corner square, we can select the other square in 1 ways.

Therefore, for this cases, favourable cases= 4*1=4

Case 2: The squares on the edges.

There are 24 squares on the edges. For each such square, we can select the other square in 2 ways.

Therefore, for this cases, favourable cases= 24*2=48

Case 3: The inner squares

There are 36 inner squares. For each such square, we can select the other square in 4 ways.

Therefore, for this cases, favourable cases= 36*4=144

On adding these three cases we get 4+48+144=196

But this is not the final answer.

We have counted every case twice.

So total favourable cases= 196/2=98

The probability= 98/2016.

Key Learning:

Remember the formulas to calculate the number of squares and rectangles in a m*n board. It can be directly applied to questions on chessboard.

Suggested Reading :
Learn the concept of probability
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