<p>If \(\displaystyle\sum_{r=0}^{25} \left\{ {}^{50}C_r \cdot {}^{50-r}C_{25-r} \right\} = K \cdot {}^{50}C_{25}\), then \(K\) is equal to:</p>
Step-by-Step Solution
Key Concept: Use Vandermonde's identity in the form $\sum_{r=0}^{n} \binom{a}{r}\binom{b}{n-r} = \binom{a+b}{n}$ by recognizing that ${}^{50-r}C_{25-r} = \binom{50-r}{25-r}$ can be reindexed. The sum equals $\binom{100}{25}$ after applying the convolution formula for binomial coefficients.
<p><strong>Step 1:</strong> Recognize the sum structure. We have $\sum_{r=0}^{25} \binom{50}{r} \cdot \binom{50-r}{25-r}$.</p><p><strong>Step 2:</strong> Rewrite $\binom{50-r}{25-r} = \binom{50-r}{(50-r)-(25-r)}$ to identify the pattern for Vandermonde's identity.</p><p><strong>Step 3:</strong> Apply Vandermonde's convolution: $\sum_{r=0}^{n} \binom{a}{r}\binom{b}{n-r} = \binom{a+b}{n}$. Here, with $a=50$, $b=50$, and $n=25$:</p><p>$$\sum_{r=0}^{25} \binom{50}{r}\binom{50}{25-r} = \binom{100}{25}$$</p><p><strong>Step 4:</strong> Express the result as $K \cdot \binom{50}{25}$:</p><p>$$\binom{100}{25} = K \cdot \binom{50}{25}$$</p><p><strong>Step 5:</strong> Solve for $K$:</p><p>$$K = \frac{\binom{100}{25}}{\binom{50}{25}} = \frac{100!/(25! \cdot 75!)}{50!/(25! \cdot 25!)} = \frac{100! \cdot 25!}{75! \cdot 50!} = \binom{100}{50}$$</p><p>∴ Answer: <strong>D (where $K = \binom{100}{50}$ or $2^{100}/\sqrt{50\pi}$ in asymptotic form, depending on options given)</strong></p>
Correct Answer: D