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

=> 1² +2² = 5

So in a 8*8 chessboard, n=8
=> No. of squares = 1² + 2² + 3²+.....+ 8²
= [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 1³ + 2³ + 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 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:

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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