SOLVE AND VERIFY
a. (n-2)! / (n-3)! = 5
b. (n+2)! / n! = 42
c. (n-3)! / 2!(n-4)! = 17
d. nP2 = 12
+2441 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.
\(\frac{(n-2)(n-3)!}{(n-3)!}=5\)
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.
\(\frac{(n+2)(n+1)n!}{n!}=42\)
This time, \(n!\) was the common factor!
\((n+2)(n+1)=42\)
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.
