1. **State the problem:** Simplify the equation $$\frac{65 n!}{(n - 5)! 5!} = \frac{2 (n - 1) (n + 1)!}{(n - 5)! 6!}$$ and solve for $n$. 2. **Rewrite the equation:** Since $(n - 5)!$ appears in both denominators, multiply both sides by $(n - 5)!$ to eliminate it: $$\frac{65 n!}{5!} = \frac{2 (n - 1) (n + 1)!}{6!}$$ 3. **Recall factorial values:** - $5! = 120$ - $6! = 720$ 4. **Substitute factorial values:** $$\frac{65 n!}{120} = \frac{2 (n - 1) (n + 1)!}{720}$$ 5. **Multiply both sides by 720 to clear denominators:** $$720 \times \frac{65 n!}{120} = 2 (n - 1) (n + 1)!$$ 6. **Simplify left side:** $$720 \times \frac{65}{120} n! = 2 (n - 1) (n + 1)!$$ Calculate $720 \times \frac{65}{120} = 6 \times 65 = 390$: $$390 n! = 2 (n - 1) (n + 1)!$$ 7. **Express $(n + 1)!$ in terms of $n!$:** $$(n + 1)! = (n + 1) n!$$ 8. **Substitute into equation:** $$390 n! = 2 (n - 1) (n + 1) n!$$ 9. **Divide both sides by $n!$ (assuming $n! \neq 0$):** $$390 = 2 (n - 1) (n + 1)$$ 10. **Simplify right side:** $$(n - 1)(n + 1) = n^2 - 1$$ So, $$390 = 2 (n^2 - 1)$$ 11. **Divide both sides by 2:** $$195 = n^2 - 1$$ 12. **Add 1 to both sides:** $$196 = n^2$$ 13. **Take the square root:** $$n = \pm 14$$ 14. **Interpretation:** Since factorials are defined for non-negative integers and $n - 5$ must be non-negative (to avoid factorial of negative), $n \geq 5$. Therefore, $n = 14$ is the valid solution. **Final answer:** $$\boxed{14}$$