SOLUTION: Find n. nP4=3(nP3) (Linear Permutations)

Question 960531: Find n. nP4=3(nP3) (Linear Permutations)
Answer by harpazo(655) (Show Source): You can put this solution on YOUR website!

P(a, b) is also written in the form of aPb = a! / (a - b)!
Here we have nP4 = 3(nP3)
n! / (n - 4)! = 3 * n! / (n - 3)!
cancelling n! we have :
1 / (n - 4)! = 3 / (n - 3)!
in fact (n - 3)! is (n - 3) * (n - 3 - 1) * (n - 3 - 2)! = (n - 3)(n - 4)(n - 5)! = ...
=> we expand it to get the same term we have on the right side :
1 / (n - 4)! = 3 / [(n - 3) * (n - 4)!]
cancelling (x - 4)! we have :
1/1 = 3/(n - 3)
n - 3 = 3
n = 6
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