How many ways can 5 friends seat themselves around a round table

A probcapability problem: In how many various means can 5 civilization sit around a round table? Is the symmeattempt of the table important?

Answer:If the symmeattempt of the table is not taken into account the variety of possibilities is 5! = 120. In this case it would certainly be the same as ordering civilization on a line. However before if rotation symmetry is taken into account, there are 5 means for people to sit at the table which are simply rotations of each other. So utilizing symmetry the answer is 24.

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So I understand why it"s 5! for the first component. Order matters so it"s a permutation without replacement, so $fracn!(n-r)!$ = $frac5!(5-5)!$ = 5! = 120

But what about the last part, is there any kind of method to explain that as a permutation/combination? How specifically is 24 calculated? (it appears to be $frac5!5$, but I do not see why)

Thank you in advancement for any explanations!

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edited Mar 2 "17 at 9:41

Ferdi
asked Mar 2 "17 at 6:40

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If you count it as you initially did, $5!$, you are over counting. For eincredibly plan, say, (May, Paul, Tina, Fran, Bob), tright here are $5$ arrangements counted within $5!$ that are inditinguishable in the circular setting. For instance, (Paul, Tina, Fran, Bob, May) or (Tina, Fran, Bob, May, Paul), (Fran, Bob, May, Paul, Tina) and (Bob, May, Paul, Tina, Fran) all considered similar to the initial (May, Paul, Tina, Fran, Bob). These are cyclic permutations.

They are counted as $(n-1)!$ Instead of $n!$

This is the same as in your case adjusting the overcounted $5!$, by separating by $5$:

$$frac5!5=4!=24.$$

The premise is that you are sindicate rotating the confront of a clock, leaving the family member positions unadjusted. Tright here is no exterior referral, just who is sitting to the left and to the ideal of each perchild.

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edited Mar 2 "17 at 8:04
answered Mar 2 "17 at 7:51

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