Show that $\binom{n}{k} < \binom{n}{k+1}$ if and only if $k < \frac{n-1}{2}$ and then use this to deduce that the maximum of $\binom{n}{k}$ for $k=0,1,\dots,n$ is $\binom{n}{\lfloor n/2\rfloor}$.
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Martin Sleziak
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user161853
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Related: http://math.stackexchange.com/questions/722952/how-do-you-prove-n-choose-k-is-maximum-when-k-is-lceil-frac-n2-rceil – Martin Sleziak Jul 06 '14 at 06:53
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Hint: If you're able to use the formula $\binom{n}{k}=\frac{n!}{k! (n-k)!}$, then it's just arithmetic: $$\binom{n}{k}=\frac{n!}{k! (n-k)!}= \frac{k+1}{n-k}\frac{n!}{(k+1)! (n-k-1)!}=\underbrace{\frac{k+1}{n-k}}_{\substack{\text{when is this} \\ \text{$<1$?}}} \binom{n}{k+1}.$$
For the second part, we can use that $\binom{n}{k}=\binom{n}{n-k}$.
Rebecca J. Stones
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I get up to part where you substituted (k+1), but where did (k+1)/(n-k) come from? – user161853 Jul 07 '14 at 02:15
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We combine $\frac{1}{k!}=\frac{k+1}{(k+1)!}$ and $\frac{1}{(n-k)!}=\frac{1}{(n-k)(n-k-1)!}$, to get $\frac{1}{k!(n-k)!}=\frac{k+1}{n-k}\frac{1}{(k+1)!(n-k-1)!}$. – Rebecca J. Stones Jul 07 '14 at 02:18
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Why is 1/k! = (k+1)/(k+1)! and not just 1/(k+1)! ? similarly, shouldn't 1/ (n-k)! = 1/ (n-k-1)! – user161853 Jul 07 '14 at 02:47
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By definition of a factorial, $(k+1)!=(k+1) \times k!$; after that, we just rearrange to get $\frac{1}{k!}=\frac{k+1}{(k+1)!}$. – Rebecca J. Stones Jul 07 '14 at 03:06
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We have $(k+1)!=(k+1) \times k!$. We divide both sides by $k!$ to obtain $\frac{(k+1)!}{k!}=k+1$, then divide both sides by $(k+1)!$ to obtain $\frac{1}{k!}=\frac{k+1}{(k+1)!}$. – Rebecca J. Stones Jul 07 '14 at 03:28
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I like using
$${n\choose k}={n-1\choose k}+{n-1\choose k-1}$$
and induction. Then you check the base case.
The pictoral representation is Pascal's triangle, you are adding bigger numbers when you go farther along (up to the line of symmetry) so the values get larger. You even get the "only if" direction along the way thanks to the symmetry in the induction.
Adam Hughes
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