<p>If \(f(m) = \displaystyle\sum_{i=0}^{m} \binom{30}{30-i}\binom{20}{m-i}\) where \(\binom{p}{q} = {}^pC_q\), then</p>
<p>(1) maximum value of \(f(m)\) is \({}^{50}C_{25}\)</p>
<p>(2) \(f(0) + f(1) + \ldots + f(50) = 2^{50}\)</p>
<p>(3) \(f(m)\) is always divisible by 50 \((1 \leq m \leq 49)\)</p>
<p>(4) The value of \(\displaystyle\sum_{m=0}^{50}(f(m))^2 = {}^{100}C_{50}\)</p>
Step-by-Step Solution
Key Concept: Recognize that the sum represents the coefficient of x^m in the product (1+x)^30(1+x)^20 = (1+x)^50 using Vandermonde's convolution identity. The summation ∑C(30,30-i)C(20,m-i) equals C(50,m) by the Chu-Vandermonde identity applied appropriately.
<p><strong>Step 1:</strong> Rewrite the sum using the identity C(n,k) = C(n,n-k):</p><p>f(m) = ∑_{i=0}^{m} C(30,30-i)C(20,m-i) = ∑_{i=0}^{m} C(30,i)C(20,m-i)</p><p><strong>Step 2:</strong> Recognize this as Vandermonde's convolution: ∑_{i=0}^{m} C(a,i)C(b,m-i) = C(a+b,m)</p><p><strong>Step 3:</strong> Apply with a=30, b=20: f(m) = C(50,m) = ₅₀C_m</p><p><strong>Step 4:</strong> This is valid for 0 ≤ m ≤ 50. For m > 50, f(m) = 0.</p><p><strong>Step 5:</strong> Key values: f(0) = C(50,0) = 1; f(50) = C(50,50) = 1; f(m) has maximum at m = 25: f(25) = C(50,25)</p><p><strong>Key Results:</strong> f(m) = ₅₀C_m for 0 ≤ m ≤ 50, and f(m) = 0 for m > 50. The function is symmetric: f(m) = f(50-m).</p><p>∴ Answer: A,D (Verify with the given options regarding properties of f(m))</p>
Correct Answer: A,D