Often times in CAT exam, questions are asked on "How Many Squares Are There In A Chessboard?" or "How Many Rectangles Are There In A 8*8 Chess-board Which Are Not Squares?", etc. These questions though look time confusing or difficult to some but are not if approached with right concept. Let's have a look at the same!

In a **2*2** chessboard, there are 5 squares (4 small squares, 1 big square)

=> 1^{2} +2^{2} = 5

So in a **8*8 chessboard**, n=8

=> No. of squares = 1^{2} + 2^{2 }+ 3^{2}+.....+ 8^{2}

= [n*(n+1)*(2n+1)]/6 (summation formula)

= **204**

Also, **number of white and black squares in chessboard:** 204/2 => 102 each

**Rectangles:**

For an **n*n** chessboard, there are 1^{3} + 2^{3 }+ 3^{3}+.....+ n^{3} rectangles.

So for an **8*8** chessboard, there are, rectangles = 1^{3} + 2^{3}+....+ 8^{3}

=> no. of rectangles = **[{n ^{2}}*{(n+1)^{2}}]/4**

n=8, we get no. of rectangles = **1296**

****No. of rectangles that are not squares** in an 8*8 chessboard

=> 1296 - 204

= **1092**

Likewise, it goes for 7*7, 6*6, 5*5, 4*4 or any other n*n matrix.

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**Chess Arrangement Problems**

**Independence problems**

Given a certain chess piece (queen, rook, bishop, knight or king) find the maximum number of such pieces, which can be placed on a chess board so that none of the pieces attack each other.

The maximum number of independent kings on an 8×8 chessboard is 16, queens - 8, rooks - 8, bishops - 14, knights - 32.

**Domination problems**

In these problems it is requested to find a minimum number of pieces of the given kind and place them on a chess board in such a way, that all free squares of the board are attacked by at least one piece.

The minimal number of dominating kings is 9, queens - 5, rooks - 8, bishops - 8, knights - 12.

** **

**What is the probability of getting two squares having a common side on a 8*8 chess board?**

**Method 1**

No. of vertical lines in a chess board (not counting the external boundaries) is 7.

Each such vertical line is divided into 8 smaller parts.

Total no of such vertical smaller parts (vertical edges of small squares) = 8*7 => 56

Similarly, no of horizontal smaller edges = 56

total number of edges = 112.

Every single edge gives a unique combination of selecting squares that have a common edge.

Total number of ways of selecting two squares = 64C2.

Hence, probability = 112/64C2 = 0.0555

**Method 2**

There are three types of squares in a chess board.

Type 1: The four corners.

Type 2: The squares at the edges but not the corners = 24

Type 3: The squares which are not in the corners nor at the edge = 36

Type 1: For each square there are two possibilities. Total = 4*2 => 8

Type 2: For each square there are three possibilities. Total = 24*3 => 72

Type 3: For each square there are four possibilities. Total = 36*4 => 144

No of ways 1^{st} square can be chosen = 64 ways.

no of ways 2^{nd} one can be chosen = 63 ways.

Total = 64*63

**Required Probability =** 224/(64*63) = **0.055**

** **

**If two squares are chosen at random on a chessboard, what is the probability that they have exactly one corner in common?**

There are 3 types of squares, a different number of each type, and each with a different number of squares they share exactly one corner with:

- Corners - 4 of them-shares exactly 1 corner with 1 square.
- Edges - 24 of them-shares exactly 1 corner with 2 squares.
- Other - 36 of them-shares exactly 1 corner with 4 squares.

Type 1: 4*1=4

Type 2: 24*2=48

Type 3: 36*4=144

Total = 196

Total Ways of choosing any 2 squares one after another = 64*63

Hence, required probability = 196/(64*63) = 0.486

**Method 3
**

In the first and second row, there are 14 ways of selecting two (1 x 1) squares that have only one common corner.

Rows 2 and 3 will similarly have 14 squares that have only one common corner.

We can pick 7 such pairs of rows in a chess board viz., (1, 2), (2, 3), (3, 4), (4, 5), (5, 6), (6, 7), and (7,8)

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