To simplify the given expression:

$\frac{(n-3)!}{(n-2)!}$

Steps:

1. Recall the factorial property: For any integer $k$, $k! = k \cdot (k-1)!$

2. Expand the denominator $(n-2)!$ in terms of $(n-3)!$: $(n-2)! = (n-2) \cdot (n-3)!$

3. Substitute this expansion into the denominator: $\frac{(n-3)!}{(n-2)!} = \frac{(n-3)!}{(n-2) \cdot (n-3)!}$

4. Cancel $(n-3)!$ in the numerator and denominator: $\frac{(n-3)!}{(n-2) \cdot (n-3)!} = \frac{1}{n-2}$

Final Simplified Expression:

$\frac{1}{n-2}$

Let me know if you'd like further clarification!

Follow-up questions:

1. What is the general rule for simplifying factorial expressions?
2. How can factorial expressions be used in combinations or permutations?
3. What happens if $n = 2$ in this expression?
4. Can you generalize a similar simplification for $\frac{(n-k)!}{(n-m)!}$?
5. How would you compute this expression for specific values of $n$?

Tip:

When simplifying factorial expressions, always expand the larger factorial to include the smaller one for easier cancellation.