# Distinguishable Objects in a Circular Arrangement

AOPS Math Jam

If you look at #9: **Please CTRL:F -> ** this: *"Ten chairs are arranged in a circle. Find the number of subsets of this set of chairs that contain at least three adjacent chairs. *"

As you see, two of the cases are just cyclic shifts of each other.

I do not understand, why they are considered different cases?

• The table is round, but the chairs are distinguishable. Once you fix chair 1 to be at the top, you no longer have to worry about cyclic shifts.
– 6005
Aug 9, 2015 at 20:15
• As DPatrick says: "The circularness of the table makes it a bit tricky. How do we account for the circularness? One way is to focus on a particular chair -- say the chair at the top of the table in the picture below:"
– 6005
Aug 9, 2015 at 20:15
• @6005, Okay that is helpful so far. So he has fixed a chair, so there are no more cyclic shifts basically? Aug 9, 2015 at 20:19
• Yes, since the chairs are labeled, if you cyclically shift them, the fixed chair (chair 1) will now be in a different place. So that is a different subset. The cyclic shift is really a red herring, since the chairs are distinguishable. See Rolf Hoyer's answer.
– 6005
Aug 9, 2015 at 20:21
• @6005, Wait a second. So If we had, take from the one left to the top, the top, and one right to the top as. $\{1, 2, 3\}$ then we take the second case and cyclic shift, is the subset $\{1, 2, 3 \}$ which we will have count as the same "subset" as the first one? (just the same arrangement cycled one to the left)? Aug 9, 2015 at 20:50

Since the chairs are distinguishable, cyclic shifts very well may yield different subsets. For instance, you want to treat $\{1,2,3\}$ and $\{2,3,4\}$ as distinct subsets of $\{1,2,3,4,5,6,7,8,9,10\}$.
• I do not understand this actually at all. Take 4 chairs for even more simplicity. $ABCD$ I know the cyclic shifts in the same order will not change anything, but there is a different in: $ABCD$ and $ABDC$ Aug 10, 2015 at 9:54