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= 8^{2}+7^{2}+6^{2}+...+2^{2}+1^{2}= 204

Can you see a pattern?

In a 8*8 chessboard, the total number of squares is ∑8^{2}

We can generalize this in the following way:

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 = ^{9}C_{2} * ^{9}C_{2} = 36*36=1296.

We can generalize this in the following way:

**Total number of rectangles in a n*n chessboard will be= ^{n+1}C_{2}*^{n+1}C_{2} **

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

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+1}C_{2}*^{n+1}C_{2}**

(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.

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**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 1^{st} rook= ^{64}C_{1}

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= ^{64}C_{2}=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= ^{64}C_{2}= 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.

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.

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