1. **State the problem:** Simplify the expression $$\frac{(n-k)(n-1)! + k(n-1)!}{k!(n-k)!}$$. 2. **Factor the numerator:** Notice that both terms in the numerator share a common factor of $(n-1)!$. $$\frac{(n-k)(n-1)! + k(n-1)!}{k!(n-k)!} = \frac{\cancel{(n-1)!}((n-k) + k)}{k!(n-k)!}$$ 3. **Simplify inside the parentheses:** $$(n-k) + k = n - k + k = n$$ 4. **Rewrite the expression:** $$\frac{\cancel{(n-1)!} \cdot n}{k!(n-k)!}$$ 5. **Recall the factorial relation:** $$n! = n \times (n-1)!$$ 6. **Substitute $n (n-1)!$ with $n!$:** $$\frac{n!}{k!(n-k)!}$$ 7. **Recognize the binomial coefficient:** The expression $$\frac{n!}{k!(n-k)!}$$ is the formula for the binomial coefficient $\binom{n}{k}$. **Final answer:** $$\boxed{\binom{n}{k}}$$