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Question:

Prove ${2n \choose r+1}$ is maximum for $r=n$

My Efforts:

$$\begin{align} {2n \choose r+1} > {2n \choose r} & \Leftrightarrow \frac{(2n)!}{(r+1)!(2n-r-1)!} > \frac{(2n)!}{r!(2n-r)!}\\ &\Leftrightarrow\frac{(2n-r)!}{(2n-r-1)!}>\frac{(r+1)!}{r!} \\ &\Leftrightarrow 2n-r > r+1 \\ &\Leftrightarrow ,r<n-\frac{1}{2} \\ \end {align}$$

I don't know how to move ahead.

P.S. my teacher gave me clue to begin with $ {2n \choose r+1} > {2n \choose r} $

Freddy
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  • So that inequality is valid as long as $r<n$, and then it reverses at $r=n$ onwards. What does that tell you about which term is largest? – Macavity Sep 15 '14 at 16:20
  • Alternatively, solve that inequality for $r$ and see what you get. – rogerl Sep 15 '14 at 16:22
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    http://math.stackexchange.com/questions/722952/how-do-you-prove-n-choose-k-is-maximum-when-k-is-lceil-frac-n2-rceil – lab bhattacharjee Sep 15 '14 at 16:40
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    Is the problem statement correct? Looks to me that the correct statement should be $\binom{2n}{r}$ is maximum when $r=n$. That is, in fact, what the answer below is proving. – Rick Goldstein Sep 15 '14 at 17:26
  • @RickGoldstein I think for n=r is equivalent to when n=r. If it is not than answer given is incorrect! – Freddy Sep 15 '14 at 17:33
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    @hey I don't understand what you're saying. Consider $n=1$. Then the binomial coefficients are $\binom{2}{0} = 1$, $\binom{2}{1} = 2$, and $\binom{2}{2} = 1$. The maximum value is the middle one, which, in your formula would be $r=0$, not $r=n=1$. – Rick Goldstein Sep 15 '14 at 17:43
  • @RickGoldstein I am learning combination for first time, even I am not getting it. I don't know whether what I have done is correct. I just moved ahead with my teacher's clue. So it's better tomorrow i will ask my teacher to help me and then i will post it as answer. – Freddy Sep 15 '14 at 17:55
  • You can't prove something that isn't true. Not if you're a mathematician, anyway. – TonyK Sep 15 '14 at 22:25

1 Answers1

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$\binom{2n}{r}\geq\binom{2n}{r-1}\Leftrightarrow \frac{2n-(r-1)}{r!(2n-(r-1))!}\geq\frac{r}{r!(2n-(r-1))!}$

$\Leftrightarrow 2n-r+1\geq r\Leftrightarrow 2r\leq2n+1$

Let's do it the other way now.

$\binom{2n}{r}\geq\binom{2n}{r+1}\Leftrightarrow \frac{r+1}{(r+1)!(2n-r)!}\geq\frac{2n-r}{(r+1)!(2n-r)!}$

$\Leftrightarrow 2r\geq 2n-1$

Therefore if $r$ is the maximum then $2n-1\leq 2r\leq2n+1$

Therefore $2r=2n$. Hence $r=n$

PS : For writing $\binom{a}{b}$ in $\LaTeX$, use \binom{a}{b} rather than {a\choose b}

Hippalectryon
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    I had spend around 15 minutes to figure out, how to write it! Thanks, that will also equally helpful :) – Freddy Sep 15 '14 at 16:43
  • @hey If that was helpful, you can upvote and click the accept answer button :) – Hippalectryon Sep 15 '14 at 16:47
  • +1 This is a good proof, but note my comment above. You have proven that $\binom{2n}{r}$ is maximal when $r=n$, not that $\binom{2n}{r+1}$ is maximal. I think the OP is a little confused, and the question statement need editing. – Rick Goldstein Sep 15 '14 at 18:00
  • @RickGoldstein Thanks, I hadn't seen that. I edited it. – Hippalectryon Sep 15 '14 at 18:38
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    @Hippalectryon - I think you misunderstood me. Your original proof is correct. Your edited proof is also correct, but in neither case have you proven the statement of the original question, because that question is misstated. Also, your statement near the end that $\binom{2n}{n}=\binom{2n}{n-1}$ is not correct--substitute $n=1$ to verify. Think about it this way: a binomial expansion of order $2n$ has $2n+1$ terms, with the largest binomial coefficient being the middle one, i.e. $\binom{2n}{n}$. – Rick Goldstein Sep 15 '14 at 21:19