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How do I prove:$$\left(\begin{array}{l}n \\ 0\end{array}\right)<\left(\begin{array}{l}n \\ 1\end{array}\right)<\cdots<\left(\begin{array}{c}n \\ \lfloor n / 2\rfloor\end{array}\right)=\left(\begin{array}{c}n \\ \lceil n / 2\rceil\end{array}\right)>\cdots>\left(\begin{array}{c}n \\ n-1\end{array}\right)>\left(\begin{array}{l}n \\ n\end{array}\right)$$ I tried using the definition of the binomial coeffient. First of all, I split it up into to parts:

$(1)$ Prove: $\begin{pmatrix} n\\m \end{pmatrix}<\begin{pmatrix} n\\k \end{pmatrix}$ for $k>m\in \{1,2,...,\lfloor n/2 \rfloor -1\}$

$(2)$ Prove: $\begin{pmatrix} n\\m \end{pmatrix}<\begin{pmatrix} n\\k \end{pmatrix}$ for $k<m\in\{n,n-1,...,\lceil n/2\rceil +1\}$

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  • $\begingroup$ Hint: Consider $\frac{n\choose k}{n\choose{k+1}}$. $\endgroup$
    – Bonnaduck
    Oct 31, 2021 at 14:42

2 Answers 2

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$$\frac{n+1}{k+1}{n\choose k}={n+1\choose k+1}={n\choose k+1}+{n\choose k}$$ $${n\choose k+1}=\frac{n-k}{k+1}{n\choose k}$$ If $\displaystyle{n\choose k}<{n\choose k+1}$ then $$\implies \frac{n-k}{k+1}>1 \\\implies k<\frac{n-1}2\le\bigg\lfloor\frac n2\bigg\rfloor$$

In the same way we can prove that $\displaystyle{n\choose k}>{n\choose k+1}$ only if $\displaystyle k>\bigg\lceil\frac n2\bigg\rceil$.

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Let me give you an hint by which you can prove it by yourself. Now also let, $$C_{n}=\frac{1}{n+1}\binom{2n}{n}$$ $\mathbf{CLAIM}:-$ $$C_{n}=\binom{2n}{n}-\binom{2n}{n+1}$$ $\mathbf{PROOF}:-$ $$\binom{2n}{n+1}=\frac{(2n)!}{(n+1)!(n-1)!}=\frac{n}{n+1}\frac{(2n)!}{(n!)^{2}}$$ $$\binom{2n}{n+1}=\frac{n}{n+1}\binom{2n}{n}$$ So $$\binom{2n}{n}-\binom{2n}{n+1}=\frac{1}{n+1}\binom{2n}{n}=C_{n}$$ Hence proved the claim. Now note that $C_n\ge1$ so $\binom{2n}{n}>\binom{2n}{n+1}$. I think this is enough for you to proceed and prove your required Inequality. The coefficients $C_{n}$ that I mentioned are called as Catalan's number. Here's a link if you are interested https://en.m.wikipedia.org/wiki/Catalan_number

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