a.   (n-2)! / (n-3)! = 5


b.   (n+2)! / n! = 42


c.   (n-3)! / 2!(n-4)! = 17


d.   nP2 = 12

 May 25, 2020

I will do the first two problems and then I will let you apply the techniques I demonstrated for the rest of them.


a) \(\frac{(n-2)!}{(n-3)!}=5\)


The key to this question is to manipulate the left-hand side of the equation by "expanding" one of the factorials until a major cancellation occurs. Here is the way I would approach it.




This manipulation is very helpful as it is now clear that \((n-3)!\) is a factor of both the numerator and the denominator. We can cancel this common factor to simplify the equation significantly.


\(n-2=5\\ n=7\)


Okay, let's apply this approach to the next one, too.


b) \(\frac{(n+2)!}{n!}=42\)


The procedure is the same: Expand one of the factorial terms until a major cancellation occurs.



This time, \(n!\) was the common factor!




This is a fairly standard quadratic. Let's solve it now.


\(n^2+n+2n+2=42\\ n^2+3n-40=0\\\)


This quadratic happens to be factorable. 


\((n-5)(n+8)=0\\ n=5\text{ or }n=-8\)


Always quickly check to make sure that your solutions are valid. Here, \(n=-8\) would be considered an extraneous solution because the input of a factorial function should always be nonnegative.

 May 25, 2020
edited by TheXSquaredFactor  May 25, 2020

14 Online Users