Without computing directly, how to show that $\dfrac{\binom{100}{50}}{2^{100}}<0.1$ easily?
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1Your question is equivalent to $\prod_{i=1}^{50} \frac {2i-1}{2i} < \frac {1}{10}$. – Calvin Lin Mar 12 '13 at 02:45
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You may use the estimate
$$\frac{1}{2^{2 n}} \binom{2 n}{n} \sim \frac{1}{\sqrt{\pi n}}$$
as $n \rightarrow \infty$. Then
$$\frac{\binom{100}{50}}{2^{100}} \approx \frac{1}{\sqrt{50 \pi}} < 0.1$$
This is derived using Stirling's Approximation. The (relative) error in the approximation is $1/(8 n)$, or about 1 part in 400 at $n=50$.
Ron Gordon
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