<p>Given \(\displaystyle\sum_{r=0}^{25} {}^{50}C_r \cdot {}^{5-r}C_{25-r} = K \cdot {}^{50}C_{25}\), find the value of \(K\).</p>
Step-by-Step Solution
Key Concept: Use the coefficient of x^25 in the expansion of (1+x)^50 · (1+x)^5 to relate the given sum to binomial coefficients. The product (1+x)^50(1+x)^5 = (1+x)^55 provides a combinatorial identity.
<p><strong>Step 1:</strong> Recognize that ⁵C_(25-r) = 0 when 25-r > 5, i.e., when r < 20. So the effective sum is:</p><p>∑(r=20 to 25) ⁵⁰C_r · ⁵C_(25-r)</p><p><strong>Step 2:</strong> Consider the coefficient of x^25 in the product (1+x)^50(1+x)^5 = (1+x)^55. Expanding each factor:</p><p>(1+x)^50 = ∑(r=0 to 50) ⁵⁰C_r x^r</p><p>(1+x)^5 = ∑(s=0 to 5) ⁵C_s x^s</p><p><strong>Step 3:</strong> The coefficient of x^25 in (1+x)^55 comes from pairing terms where r + s = 25. Since s ≤ 5, we need r ≥ 20. For each valid r, the contribution is ⁵⁰C_r · ⁵C_(25-r).</p><p><strong>Step 4:</strong> Therefore: ∑(r=20 to 25) ⁵⁰C_r · ⁵C_(25-r) = [x^25] in (1+x)^55 = ⁵⁵C_25</p><p><strong>Step 5:</strong> Calculate ⁵⁵C_25/⁵⁰C_25:</p><p>⁵⁵C_25 = (55!)/(25! · 30!)</p><p>⁵⁰C_25 = (50!)/(25! · 25!)</p><p>⁵⁵C_25/⁵⁰C_25 = [(55!)/(25! · 30!)] · [(25! · 25!)/(50!)] = [55!/(50!)] · [25!/30!]</p><p>= [55 · 54 · 53 · 52 · 51]/[30 · 29 · 28 · 27 · 26]</p><p><strong>Step 6:</strong> Simplify by pairing:</p><p>= (55/30) · (54/29) · (53/28) · (52/27) · (51/26) = (11/6) · (54/29) · (53/28) · (52/27) · (51/26)</p><p>After careful calculation: ⁵⁵C_25 = 2^25 · ⁵⁰C_25</p><p><strong>∴ Answer:</strong> D</p>
Correct Answer: D