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Ireland counting and probability
Problem
For , an n-spinner is a “fidget spinner” with identical arms and an angle of degrees between pairs of adjacent arms. For instance, a -spinner and a -spinner are illustrated below: 
We wish to place each of the numbers on the arms of an -spinner, with numbers placed on each arm, all equally spaced in a row. Some possibilities for are illustrated below.
Since the first two can be obtained by rotation from each other, they are considered to be the same, and so we count the above as only two different placements. (The third is genuinely different because the “5” is closer to the middle than it is in the other two.) Counting in this manner, how many different placements do we obtain for given and ?
We wish to place each of the numbers on the arms of an -spinner, with numbers placed on each arm, all equally spaced in a row. Some possibilities for are illustrated below.
Since the first two can be obtained by rotation from each other, they are considered to be the same, and so we count the above as only two different placements. (The third is genuinely different because the “5” is closer to the middle than it is in the other two.) Counting in this manner, how many different placements do we obtain for given and ?
Solution
First, place the spinner so that the arm containing the number is at the top. There, it can be put in any of the positions on that arm. This orients the -spinner and breaks the symmetry. The other numbers can now be placed in any other positions, so the answer is .
Alternatively, we place all numbers as we wish, giving permutations. We then divide by since each placement can be rotated to different equivalent placements. Since , the answers are consistent.
Alternatively, we place all numbers as we wish, giving permutations. We then divide by since each placement can be rotated to different equivalent placements. Since , the answers are consistent.
Final answer
m(nm - 1)! = (nm)!/n
Techniques
Enumeration with symmetryCounting two ways