AIME Problem 8

Let's divide the board up into 13 disjoint regions: the central square, and 12 buckets , where each bucket is a region of 4 squares that is invariant on rotation. For example, the four corners of the board are one bucket.

Then we can separate the possible locations of yellow squares into three cases:

• Case 1: One square in each of two distinct buckets
• Case 2: Both squares in one bucket
• Case 3: The central square, and one bucket.

Solving each of these cases separately:

Case 1. There are $12 \choose 2$ ways of choosing the two buckets. If we fix the location of one yellow square in one bucket then we see that there are $4$ ways of placing the other yellow square.

Case 2. There are $12$ ways of choosing the bucket; and then $2$ ways of placing the yellow squares within that bucket (either adjacent or opposite).

Case 3. After placing the central square, there are $12$ buckets in which the other yellow square could be placed.

Summing the three cases gives $4 \times {12 \choose 2} + 12 \times 2 + 12 = \boxed{300}$

Every pattern repeats either 4 times or 2 times,

Let the number of patterns repeating 4 times be $x$

and the numbers of pattern repeating 2 times be $y$

Then, $4x+2y=$ $\left( \begin{matrix} 49 \\ 2 \end{matrix} \right)$ ,

And $(x+y)$ will be the wanted result.

All the patterns in $y$ must be such that when rotated twice it regains it original position,

So if one color was at $(a,b)$ and the other one must be $(8-a,8-b)$ .

$a$ can be $1,2,3,4,5,6,7$ and so can $b$

So there are $7*7=49$ pairs for $y$ but the block $(4,4)$ won't be valid since the other block will be $(4,4)$ and will coincide

There $(4,4)$ won't be considered for $y$ So there are $48$ ways of $(a,b)$ existing.

But the 2 blocks are identical so $y$ is $48/(2*2)$ for $2$ ways of selecting $a$ and $2$ ways of selecting $b$ .

So $y=12$ and $4x+2y=$ $\left( \begin{matrix} 49 \\ 2 \end{matrix} \right)$ , $=1176$

$x=288$

$x+y=300$

So there are 300 different patterns in which the chessboard can be colored