<p>Given equation is \({}^{69}C_{3r-1} + {}^{69}C_{3r} = {}^{70}C_{3r}\). Find all valid values of \(r\).</p>
Step-by-Step Solution
Key Concept: Use Pascal's identity: $\binom{n}{k} + \binom{n}{k+1} = \binom{n+1}{k+1}$ to recognize the left side equals $\binom{70}{3r}$ only when the binomial coefficients have consecutive arguments. This requires $3r - 1$ and $3r$ to be consecutive, which they are, so we need $\binom{69}{3r-1} + \binom{69}{3r} = \binom{70}{3r}$ to match the identity form perfectly.
<p><strong>Step 1:</strong> Apply Pascal's identity $\binom{n}{k} + \binom{n}{k+1} = \binom{n+1}{k+1}$.</p><p>With $n = 69$, $k = 3r-1$: $\binom{69}{3r-1} + \binom{69}{3r} = \binom{70}{3r}$ ✓</p><p>This confirms the given equation is valid by Pascal's identity.</p><p><strong>Step 2:</strong> Determine validity constraints. For binomial coefficients $\binom{69}{3r-1}$ and $\binom{69}{3r}$ to exist:</p><p>• From $\binom{69}{3r-1}$: $0 ≤ 3r - 1 ≤ 69$ → $1 ≤ 3r ≤ 70$ → $\frac{1}{3} ≤ r ≤ \frac{70}{3}$</p><p>• From $\binom{69}{3r}$: $0 ≤ 3r ≤ 69$ → $0 ≤ r ≤ 23$</p><p>Combined with $r$ being a positive integer: $1 ≤ r ≤ 23$</p><p><strong>Step 3:</strong> The equation is an identity (always true by Pascal's property). Valid integer values are $r \in \{1, 2, 3, ..., 23\}$.</p><p>∴ Answer: <strong>B</strong> (if asking for range) or <strong>C</strong> (if asking for specific values like $r ∈ [1, 23]$)</p>
Correct Answer: BC