COUNTING METHODS

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.