<p>The sum \(\displaystyle\sum_{0\le i\le j\le 10} \binom{10}{C_j}\binom{j}{C_i}\) is equal to</p>
Step-by-Step Solution
Key Concept: Recognize this as counting lattice paths: each term counts pairs (i,j) where 0≤i≤j≤10, and reorganize using the Vandermonde convolution identity by fixing the inner index structure.
<p><strong>Step 1:</strong> Rewrite the sum by changing summation order. We have $$\sum_{0\le i\le j\le 10} \binom{10}{j}\binom{j}{i}$$ (assuming notation C_j means j, C_i means i).</p><p><strong>Step 2:</strong> Fix j first, then sum over valid i: $$\sum_{j=0}^{10}\binom{10}{j}\sum_{i=0}^{j}\binom{j}{i}$$</p><p><strong>Step 3:</strong> The inner sum $\sum_{i=0}^{j}\binom{j}{i} = 2^j$ by binomial theorem.</p><p><strong>Step 4:</strong> Now evaluate: $$\sum_{j=0}^{10}\binom{10}{j}\cdot 2^j = \sum_{j=0}^{10}\binom{10}{j}\cdot 2^j \cdot 1^{10-j}$$</p><p><strong>Step 5:</strong> By binomial theorem, this is $(2+1)^{10} = 3^{10}$.</p><p>∴ Answer: <strong>D</strong> (3^10)</p>
Correct Answer: D