Do You Know The Number Of Squares And Rectangles In A Chessboard

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!

Formula: Number of squares in a n*n Chessboard: 12 +22 +….+ n2  or [n*(n+1)*(2n+1)]/6

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

=> 12 +22 = 5

So in a 8*8 chessboard, n=8
=> No. of squares = 12 + 2+ 32+…..+ 82
= [n*(n+1)*(2n+1)]/6 (summation formula)
= 204

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

Number-of-squares-rectangles-in-chess-board

Rectangles:

For an n*n chessboard, there are 13 + 2+ 33+…..+ n3 rectangles.

So for an 8*8 chessboard, there are, rectangles = 13 + 23+….+ 83

=> no. of rectangles = [{n2}*{(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 1st square can be chosen = 64 ways.
no of ways 2nd 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:

  1. Corners – 4 of them-shares exactly 1 corner with 1 square.
  2. Edges – 24 of them-shares exactly 1 corner with 2 squares.
  3. 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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Comments

2 comments

Rittu Paul

You have added the no. of possible outcomes for Type 1, Type 2, Type 3 cases but in the process arent we counting each selection two times like wen we consider a particular square say 1 sharing one side with square 2, we are again counting square 2 sharing the same side with square 1 so final answer should be half of what you are claiming. Please check and correct because this post might mislead 1000 of aspirants.