Problem 6.1.26 From The Arts and Crafts of Problem Solving

Question

Can someone help me with problem 6.1.26? So far I did the binomial coefficient (49 2), which is the number of ways to select two out of 49 tiles. I had considered that every tile except the middle has a symmetrical counterpart, so I did (49-1)/2 = 24 and then BN(49 2) - 24. Then, I divided by 4 for the distinct amount of boards. This is different than the solution, but close. I have attached the solution below the question.

*BN for Binomial Coefficient.

I don't get where the +(24/12) comes from.

The Link to The Solution: http://artofproblemsolving.com/wiki/index.php?title=1996_AIME_Problems/Problem_7

Someone mentioned "If the two squares you pick are 'diametrically opposed,' i.e., they're on the exact opposite side of the board of the other one, then there aren't four rotations where these pairs show up. In this case, there are only two rotations possible. Imagine picking the square in the top left corner and in the bottom right corner: rotate that 90 degrees, and you get a new equivalent coloring; rotate that again 90 degrees, and you end up with the same thing you started with. Thus these only are only counted twice, not four times, so by dividing by four, you're under-counting them, and you need to correct this by adding 24/2 (since there are 24 pairs of these squares).", which I still cannot comprehend.

Answer 1

You have found there are

• ${49 \choose 2}=1176$ ways of choosing two points
• $\frac{49-1}{2}=24$ of these are symmetric across the centre point
• so $1152$ are not, and each one can be rotated to match three other pairs, making $\frac{1152}{4}=288$ unique pairs taking account of rotation
• the $24$ can only be rotated to match one other pair, making $\frac{24}{2}=12$ unique pairs taking account of rotation - you seem to have missed these
• the answer is therefore $288+12=300$ overall unique pairs taking account of rotation

This is $\dfrac{{49 \choose 2}-\frac{49-1}{2}}{4} + \dfrac{\frac{49-1}{2}}{2}=300$