Past exam of the graph Theory and Probabilistic Combinatorics course of the University of Oxford / 2008 / B / 5 / a / i

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- Solution i

Solution

For

k \geq 1

, monotonicity of the logarithm gives

lo g (k!) = \sum_{j = 1}^{k} lo g j \geq \int_{1}^{k} lo g t d t = k lo g k - k + 1 \geq k lo g k - k .

Thus

k! \geq (k / e)^{k}

. Bounding the numerator of the binomial coefficient by

n^{k}

yields

(k n) \leq \frac{n ^{k}}{k !} \leq (e n / k)^{k} .

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