<p>The value of \({}^{50}C_4 + \displaystyle\sum_{r=1}^{6} {}^{56-r}C_3\) is</p>
Step-by-Step Solution
Key Concept: Use the hockey stick identity: ∑C(n,r) = C(n+1,r+1). Recognize that the sum ∑_{r=1}^{6} C(56-r,3) telescopes when reindexed, and combine with the first term using this identity.
<p><strong>Step 1:</strong> Reindex the sum. Let k = 56 - r. When r goes from 1 to 6, k goes from 55 to 50:</p><p>∑_{r=1}^{6} C(56-r,3) = ∑_{k=50}^{55} C(k,3) = C(50,3) + C(51,3) + C(52,3) + C(53,3) + C(54,3) + C(55,3)</p><p><strong>Step 2:</strong> Apply the hockey stick identity repeatedly:</p><p>C(50,3) + C(51,3) = C(51,4)</p><p>C(51,4) + C(52,3) = C(52,4)</p><p>C(52,4) + C(53,3) = C(53,4)</p><p>C(53,4) + C(54,3) = C(54,4)</p><p>C(54,4) + C(55,3) = C(55,4)</p><p><strong>Step 3:</strong> Therefore: ∑_{r=1}^{6} C(56-r,3) = C(55,4)</p><p><strong>Step 4:</strong> Now combine with the first term:</p><p>C(50,4) + C(55,4)</p><p><strong>Step 5:</strong> Apply hockey stick one more time going forward from C(50,4):</p><p>C(50,4) + C(51,4) + C(52,4) + C(53,4) + C(54,4) + C(55,4) = C(56,5)</p><p><strong>Note:</strong> The expression ∑_{r=1}^{6} C(56-r,3) gives us C(51,4) + C(52,4) + C(53,4) + C(54,4) + C(55,4), which when combined with C(50,4) yields C(56,5).</p><p>∴ Answer: <strong>C(56,5)</strong></p>
Correct Answer: D