1. The problem asks to solve the combination equations: 2. First, recall the formula for combinations: $$C(n, r) = \frac{n!}{r!(n-r)!}$$ where $n!$ is the factorial of $n$. 3. For the first equation $C(n, 2) = 10$: $$C(n, 2) = \frac{n!}{2!(n-2)!} = \frac{n(n-1)}{2} = 10$$ 4. Multiply both sides by 2: $$n(n-1) = 20$$ 5. Expand and rearrange: $$n^2 - n - 20 = 0$$ 6. Solve the quadratic equation using the quadratic formula: $$n = \frac{1 \pm \sqrt{1 + 80}}{2} = \frac{1 \pm \sqrt{81}}{2} = \frac{1 \pm 9}{2}$$ 7. The two solutions are: $$n = \frac{1 + 9}{2} = 5 \quad \text{or} \quad n = \frac{1 - 9}{2} = -4$$ Since $n$ must be a positive integer, $n = 5$. 8. For the second equation $C(13, r) = 286$: $$C(13, r) = \frac{13!}{r!(13-r)!} = 286$$ 9. We need to find $r$ such that $C(13, r) = 286$. We can check values of $r$: - $C(13, 3) = \frac{13 \times 12 \times 11}{3 \times 2 \times 1} = 286$ 10. Therefore, $r = 3$. **Final answers:** $$n = 5$$ $$r = 3$$