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Prove that there are $n! - (n-1)(n-1)!$ ways to arrange $n$ objects in a circular arrangement.

I have tried algebraic proofs by equating it to $\frac{n!}{n}$ and to $(n-1)!$ but can't think of a way to prove this combinatorially.

dingus
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    What have you tried beyond $n! - (n-1)(n-1)! = \frac{n!}{n}= (n-1)! $ ? Some approaches might give the second or third expressions. You might look at small examples such as $n=3$ or $4$ to see which arrangements are equivalent. – Henry Feb 19 '24 at 08:39
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    try to think about n!/n, we need to divide by n since we are working in a circular context, try to visualize what I mean, play with n = 4. – Pat Feb 19 '24 at 08:44
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    Think about people sitting at a round table. For a given arrangement, if everyone decides to shift one chair to their right, then the arrangement remains unaltered. Based on this, give meaning to $n!/n$. Combinatoric problems often require you to think of specific scenarios, and give meaning to them. See also: https://math.stackexchange.com/questions/2387149/explanation-circular-permutation – Oscar Feb 19 '24 at 08:55

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The number of circular permutations (rotations don't matter) of $n$ objects can be thought of as linear permutations of $n$ objects with the restriction that one of those objects is fixed at a certain position. [I assume you already know this.]

The number of linear permutations of $n!$ objects with no restrictions is $n!$. The number of permutations of $n$ objects such that a certain object is fixed at any specific position is $(n-1)!$.

Say, we choose to fix the chosen object in the first position in the line. That means we must subtract from $n!$ all the permutations in which that objet is not in the the first position.There are $n-1$ positions other than the first. So that number of unwanted permutations would be $(n-1) (n-1)!$.

Subtracting it from the total $n!$, we get the number of circular permutations

$$n! - (n-1)(n-1)!$$

Haris
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