I know that if the chairs were not distinct, then the answer is $(n-1)!$. This is because in this case relative position of the persons matter and not actually where they are seated.

But since now the chairs are also distinctly colored, the first person has $n$ distinct chairs to sit. Then the second person has $(n-1)$ distinct chairs to sit and so on...

So by multiplication rule we have total $n!$ arrangements.

Is my understanding correct?

  • 2
    $\begingroup$ That's right. Distinctly coloured chairs amount to the same as numbered chairs. $\endgroup$ – true blue anil Oct 15 '17 at 15:13
  • $\begingroup$ Thanks! ${}{}{}{}{}{}{}$ $\endgroup$ – Error 404 Oct 17 '17 at 3:16
  • $\begingroup$ You're welcome ! $\endgroup$ – true blue anil Oct 17 '17 at 4:35

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