Question: If six people, designated as A,B,C,D,E,F, are seated about a round table how many different circular arrangements are possible, if arrangements are considered the (same) when one can be obtained from the other by rotation?
Note: Since I don't know how to post images here/ too lazy to post them, you will have to figure out this problem by imagining the diagram or drawing it out on your own.
My attempt: Even though this question deals with circular arrangements, we can still use the permutation formula of indistinguishable objects dealing with linear arrangements: n!/nr!
First, there are 6! ways to arrange the positions of the six letters A-F; rotating each letter one at a time to a new position of the table. When a letter, say A, returns to the position it started on, that is considered a repetition. Therefore, 6! is the total number of possible circular arrangements.
That makes the n! = 6!. So the bottom, nr!, is equal to 6, because say, 1)ABCDEF, 2)BCDEFA, 3)CDEFAB, 4) DEFABC, 5) EFABCD, 6) FABCDE, are all the same arrangements. Therefore, nr! = 6, because 6 is the number of indistinguishable objects.
Dividing 6!/6 we get 120 , which is the correct answer.
My problems with this question:
From my way of doing the question above, is it the right way to do it? Although nr! = 6, why isn't it 6! ? Shouldn't it be 6! for the bottom, since that's how the formula is stated?
Another way of doing this is (n-1)! Why is this also correct? Why is 1 subtracted from n?
- Also another way is to fix the starting position of any letter, say A. That will give 5! which is 120. What does fixing one of the letters have to do finding the possible arrangements? Really confused...