ways to arrange positive integers from 1 to 100 on a circle In how many ways can the positive integers from 1 to 100 be arranged in a circle such that the
sum of every two integers placed opposite each other is the same? (Arrangements that are rotations
of each other count as the same.) Express your answer in the form a! · b^c
.
Since the question states "arrangements that are rotations of each other count as the same" , you need to be a bit careful.

After fixing the 1-100 diameter, 

there are only 49 pairs left to be fixed and flipped, yielding $49!\cdot2^{49}$

If you flip the 1-100 diameter too, you will be duplicating previous arrangements looked at (say) clockwise.
For the sum of every opposing pair to be equal there will be only one potential partner for any given integer. How many ways can we arrange these fifty diameter pairs? Using circular permutation we know $49!$ is the number of ways we can arrange fifty items in a circle, but the pairs can be individually flipped. That means each potential flip doubles the possible configurations $2^{50}$

$49! \cdot 2^{50}$