Peter R. answered 07/28/22
Adjunct Lecturer - Math Department - Borough of Manhattan C.C.
The 1st seat can be filled 9 different ways. The 2nd seat can be filled 8 different ways (the first seat already has one of the nine people there, so can't also sit in the 2nd seat). Similarly, the 3rd seat has 7 possible people. So the number of PERMUTATIONS is 9 x 8 x 7 = 504 [or 9 nPr 3 (9 people taken 3 at a time) on your scientific calculator].
This assumes that there is a meaningful difference between seating order ABC, CBA, ACB, etc. If these are considered the same (order doesn't matter) then there would be only 84 ways to seat them (9 nCr 3) on your calculator).
Peter R.
07/28/22
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