Posted by aggie60 on October 28, 2002 at 00:27:48:
In Reply to: Re: Combinations posted by Joel on October 28, 2002 at 00:09:03:
: : 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
: n=7 (7*6*5=210)
: So there were 7 empty seats before seating the 3 new students.