Expert: Sherry Wallin - 3/20/2010

Question
((n-r)!/(n-6)!)=30

(n-5)(n-4)=30
n^2-9n-10=0
(n-10)(n+1)=0
n=10

please explain it

Answer
Pratap~
The problem is to find the value of r given all the other information (I think).
You are given (n-5)(n-4)=30 and when you multiply it out you get n^2-9n+20=30 and when you move everything to the left you get n^2-9n-10=0. (n-10)(n+1)=0 is just the factorization of the previous. n -10 = 0 implies n = 10. n+1 = 0 implies n = -1 but n canNOT equal -1 because permutations are not defined for negative integers, only on non negative integers.So if you substitute n = 10 into the first expression ((n-r)!/(n-6)!) = 30 you get:

((n-r)!/(10-6)!)=30 -> ((n-r)!/4!)=30 ->
((n-r)!=4!*30 = 4*3*2*1 *30 = 30*4*3*2*1 = 6*5*4*3*2*1 = 6!
So now you have (n-r)! = (10-r)! = 6! so what does this make r??
r has to be 4 for (10-4)! = 6!

Math Prof