Question

The number of ways in which the squares of a $8\times 8$ chess board can be painted red or blue so that each $2\times 2$ square has two red and two blue square is
A. $2^9$
B. $2^9-1$
C. $2^9-2$
D. None of these.

Answer 1

Hint: First we will draw a normal $8\times 8$ chess board which has $8$ rows and $8$ columns. This means that there are a total $64$ small squares but we have to choose a $2\times 2$ square. As we know that to color $m\times n$ chess board there are $2^m+2^n-2$ possible ways. So, we use this to find a correct answer.

Complete step-by-step answer:
First we draw a diagram of a chess board with $8$ rows and $8$ columns.

We have to paint it red or blue so that each $2\times 2$ square has two red and two blue squares.
As we know that t to color $m\times n$ chess board there are $2^m+2^n-2$ possible ways.
We have a $8\times 8$ chess board, so the possible ways to color it will be $2^8+2^8-2$
When we solve this equation we get ${2.2}^8-2$
As we know that $a^m.a^n=a^{m+n}$
So the equation becomes $2^9-2$
So the total number of ways the squares of a $8\times 8$ chess board can be painted red or blue so that each $2\times 2$ square has two red and two blue squares is $2^9-2$ .
Option C is the correct answer.

Note: The key factor in this question is that a student knows the number of rows and columns of the chess board. Also, by drawing the diagram things get clearer to students. The possibility of mistake can be in solving the equation $2^8+2^8-2$ . Some students can add the powers as $2^{16}-2$ . So keep the point in mind while solving the equation.