1. **Problem Statement:** Find the value of $n$ if $$\frac{n!}{(n-3)!} = 990.$$ 2. **Formula and Explanation:** Recall that factorial notation means the product of all whole numbers from $n$ down to 1. The expression $$\frac{n!}{(n-3)!}$$ can be simplified by expanding the factorials: $$\frac{n!}{(n-3)!} = n \times (n-1) \times (n-2)$$ This is because the $(n-3)!$ cancels out all terms from $(n-3)$ down to 1 in the numerator. 3. **Set up the equation:** $$n \times (n-1) \times (n-2) = 990$$ 4. **Solve the cubic equation by trial or factorization:** Try integer values for $n$ starting from 1 upwards: - For $n=10$: $$10 \times 9 \times 8 = 720$$ (too small) - For $n=11$: $$11 \times 10 \times 9 = 990$$ (matches!) 5. **Answer:** $$\boxed{n=11}$$ This means when $n=11$, the equation holds true.