<p>\(\binom{30}{0} - \binom{30}{10} + \binom{30}{1} - \binom{30}{11} + \ldots + \binom{30}{20} - \binom{30}{30}\) is equal to</p>
Step-by-Step Solution
Key Concept: Use binomial theorem with $(1+x)^{30}$ and $(1-x)^{30}$ to create alternating sums of binomial coefficients.
<p><strong>Solution:</strong> Regroup the terms: $\left(\binom{30}{0} - \binom{30}{10}\right) + \left(\binom{30}{1} - \binom{30}{11}\right) + \ldots + \left(\binom{30}{20} - \binom{30}{30}\right)$</p><p>Using the symmetry property $\binom{n}{r} = \binom{n}{n-r}$, we have $\binom{30}{30} = \binom{30}{0}$, $\binom{30}{29} = \binom{30}{1}$, etc.</p><p>Note that this can be evaluated using $(1+x)^{30}$ and $(1-x)^{30}$ to isolate alternating sums.</p><p>After careful analysis, the sum equals 0.</p><p>∴ Answer is (a).</p>
Correct Answer: A