Question:

<p><strong>Step 1:</strong> Start with the given equation:<br/>${}^{69}C_{3r-1} - {}^{69}C_{r^2} = {}^{69}C_{r^2-1} - {}^{69}C_{3r}$</p><p><strong>Step 2:</strong> Rearrange by moving terms:<br/>${}^{69}C_{3r-1} + {}^{69}C_{3r} = {}^{69}C_{r^2-1} + {}^{69}C_{r^2}$</p><p><strong>Step 3:</strong> Apply the symmetry property ${}^nC_r = {}^nC_{n-r}$. Note that ${}^{69}C_{3r} = {}^{69}C_{69-3r}$ and ${}^{69}C_{r^2} = {}^{69}C_{69-r^2}$.</p><p><strong>Step 4:</strong> For the equation to hold, we use the property that if ${}^{69}C_a = {}^{69}C_b$, then either $a = b$ or $a + b = 69$.</p><p><strong>Step 5:</strong> From the rearranged equation and applying symmetry, we get two possible cases:<br/><strong>Case 1:</strong> $3r - 1 = r^2 - 1$, which gives $3r = r^2$ or $r^2 - 3r = 0$, so $r(r-3) = 0$, giving $r = 0$ or $r = 3$.<br/>Check: $r = 0$ gives negative indices, so invalid. $r = 3$ is valid.</p><p><strong>Step 6:</strong> <strong>Case 2:</strong> $3r - 1 + r^2 = 69$, which gives $r^2 + 3r - 70 = 0$.<br/>Using the quadratic formula: $r = \frac{-3 \pm \sqrt{9 + 280}}{2} = \frac{-3 \pm \sqrt{289}}{2} = \frac{-3 \pm 17}{2}$<br/>So $r = \frac{14}{2} = 7$ or $r = \frac{-20}{2} = -10$.<br/>Since $r$ must be non-negative, $r = 7$ is valid.</p><p><strong>Step 7:</strong> Verify constraints for $r = 3$: All indices $3r-1 = 8$, $r^2 = 9$, $r^2-1 = 8$, $3r = 9$ satisfy $0 \leq \text{index} \leq 69$. ✓</p><p><strong>Step 8:</strong> Verify constraints for $r = 7$: All indices $3r-1 = 20$, $r^2 = 49$, $r^2-1 = 48$, $3r = 21$ satisfy $0 \leq \text{index} \leq 69$. ✓</p><p><strong>∴ Answer:</strong> r = 3 or r = 7</p>