1. **Problem statement:** Find the $n$-th derivative of the function $$y = x^{2m} e^{ax}$$ using Leibniz's rule. 2. **Recall Leibniz's rule for the $n$-th derivative of a product:** $$\frac{d^n}{dx^n}[u(x)v(x)] = \sum_{k=0}^n \binom{n}{k} u^{(k)}(x) v^{(n-k)}(x)$$ where $u^{(k)}$ and $v^{(n-k)}$ are the $k$-th and $(n-k)$-th derivatives of $u$ and $v$ respectively. 3. **Identify $u$ and $v$:** Let $$u = x^{2m}, \quad v = e^{ax}$$ 4. **Derivatives of $u$:** The $k$-th derivative of $u$ is $$u^{(k)} = \frac{d^k}{dx^k} x^{2m} = \frac{(2m)!}{(2m-k)!} x^{2m-k}$$ for $k \leq 2m$, and zero otherwise. 5. **Derivatives of $v$:** Since $v = e^{ax}$, $$v^{(n-k)} = a^{n-k} e^{ax}$$ 6. **Apply Leibniz's rule:** $$\frac{d^n}{dx^n} y = \sum_{k=0}^n \binom{n}{k} \frac{(2m)!}{(2m-k)!} x^{2m-k} a^{n-k} e^{ax}$$ 7. **Note the upper limit:** Since $u^{(k)}=0$ for $k > 2m$, the sum upper limit is $\min(n, 2m)$. 8. **Final expression:** $$\frac{d^n}{dx^n} y = e^{ax} \sum_{k=0}^{\min(n, 2m)} \binom{n}{k} \frac{(2m)!}{(2m-k)!} x^{2m-k} a^{n-k}$$ 9. **Match with given options:** This matches option B. **Answer:** B