Posted by Joel on October 28, 2002 at 00:09:03:
In Reply to: Combinations posted by Aggie60 on October 27, 2002 at 23:08:38:
: realize it is easy??? but. Three (3) new students and teacher says she has 210 seating combinations. How many actual seats are empty?
I think you mean permutations. When you are talking about "how many different seating arrangements ...", order matters. The formula for an r-permutation from a set of n elements is n!/(n-r)! Here n is unknown but you know that r is 3, so
n!/(n-3)! = 210
n*(n-1)*(n-2) = 210
So there were 7 empty seats before seating the 3 new students.
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