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I was seeing this video by Eddie woo on circular seating's, in it , he discusses the following question:

Find the number of ways to arrange 5 boys and 5 girls in a circular seating

At 0:40, he places the first person and then keeps the rest and hence total ways to do it is $ 1 \cdot 9!$ , however I can't precisely frame why we only have one way to keep the first person. So, I'm looking for the understanding on why he said the first step like this... (he laters say it is related to the fact we can rotate the circle but I didn't get that either)

tryst with freedom
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    There are 10! ways to arrange 10 objects. However, if we are seating in a circle we have decided that a rotation, i.e. move everyone moves one space to the left, keeping all of the relative positions intact, won't count those as a different arrangement. One way to think about this is to divide your 10! arrangements by the 10 rotations. Another way to think about it would be to fix the location of one person. We can always rotate the arrangement to move the designated person to his fixed location. – Doug M Feb 09 '21 at 00:35

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Imagine you have placed your 10 persons around the circle. Imagine there are K distinct ways to do this placement (i.e. K such circular configurations).

Think about the circumference as if it's a circular rope e.g.

For each such configuration (circular placement) you can cut the circumference in between any two persons. You can do this in 10 different ways for each configuration (because you have 10 gaps between the 10 people). So each such circular configuration gives rise to 10 normal (linear) permutations (you can prove this is a bijection) of the 10 people.

But we know all the distinct linear permutations are $M = 10!$

So the number K should be such that $$K \cdot 10 = M$$ i.e. $$K = \frac{10!}{10}$$

Btw, it doesn't matter if you have 5 boys and 5 girls or say 8 boys and 2 girls. When the word is about people, it is always assumed that any two people are distinguishable/distinct.

peter.petrov
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