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How can one prove that for $n \equiv 0$ mod $2$ we have

$${n \choose 0} < {n \choose 1} <\ldots< {n \choose n/2-1}<{n \choose n/2}>{n\choose n/2+1}>\ldots>{n\choose n-1}>{n\choose n}\,?$$

Can I say that for fixed $n$, the binomial coefficients $n \choose k$ increase with $k$ for $k < n/2$? If n is even (like in our case), then the central binomial coefficient $n \choose n/2$ is the largest one.

So $n \choose k+1$ is greater than, equal to, or less than $n \choose k$ according as $n-k$ is greater than, equal to, or less than $k+1$, that is according as $k$ is less than, equal to, or greater than $(n−1)/2$.

Is that correct?

Batominovski
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user1234567890
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    Consider the ratio of successive binomial coefficients. – Sri-Amirthan Theivendran Nov 04 '18 at 23:28
  • Or... look at the Pascal triangle. The number of $k$-element subsets of $n$-element set is equal to the number of its $n-k$-element subsets. So, the binomial coefficients grow up to more or less $k=n/2$ and then they fall. – szw1710 Nov 04 '18 at 23:35
  • Changed my proof. Is that O.K.? – user1234567890 Nov 04 '18 at 23:48
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    Hello. Please see whether this helps you: https://math.stackexchange.com/questions/2496836/combinatorial-proof-that-central-binomial-coefficients-are-the-largest-ones –  Nov 07 '18 at 08:03
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    It's a straightforward calculation $$\binom{n}{m-1} < \binom{n}{m} \iff \frac{n!}{(m-1)!(n-m+1)!}<\frac{n!}{m!(n-m)!} \iff \ (m-1)!(n-m+1)! > m!(n-m)! \iff (n-m+1) > m \iff n+1 > 2m \iff \ \frac{n+1}{2} > m $$ 100 bounties is a lot for this question, imho. – rtybase Nov 07 '18 at 08:56

3 Answers3

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All you need is this:

Lemma: $${n\choose k-1}<{n\choose k} \iff k\leq {n\over 2}$$

Proof: You can ask yourself:

In how many ways can we choose $k$ people, including one president, from a population of $n$ people?

First answer:

You choose $k-1$ people from $n$ people and then from the rest of them a president, so we have $${n\choose k-1}\cdot {n-k+1\choose 1}$$

Second answer:

You choose $k$ people from $n$ people and then from these selected you choose a president, so we have $${n\choose k}\cdot {k\choose 1}$$

So we have:$${n\choose k-1}\cdot (n-k+1)=k{n\choose k} $$ and from here we get $${n\choose k-1}<{n\choose k} \iff k\leq {n\over 2}$$

nonuser
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Let $n=2m$. You want to prove, that for

  1. $0 \leq k<m$ we have ${{n}\choose{k}} < {{n}\choose{k+1}}$
  2. $m\leq k< n$ we have ${{n}\choose{k}} > {{n}\choose{k+1}}$

First 1.: Let $0 \leq k<m$. Let's study the ratio

$$\frac{{{n}\choose{k+1}}}{{{n}\choose{k}}} = \frac{\frac{n!}{(k+1)!(n-(k+1))!)}}{\frac{n!}{k!(n-k)!}} = \frac{k!(n-k)!}{(k+1)!(n-k-1)!} = \frac{n-k}{k+1} > 1$$ since $n-k > n - m = n- \frac{n}{2} = \frac{n}{2} = m \geq k+1$.

Then 2.: Assume $m\leq k< n$. It goes similarly, the ratio $$\frac{{{n}\choose{k}}}{{{n}\choose{k+1}}} = \frac{k+1}{n-k} > 1 $$ because now $n-k \leq n-m = m < k+1 $.

ploosu2
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Let's try using induction on $m$ where $n=2m$. For $m=0$, there is nothing to be proven, for $m=1$, then ${2\choose 0} = {2\choose 2} = 1 < 2 = {2\choose 1}$.

To get the induction step, observe that ${n+2\choose k} = {n+1\choose k-1} + {n+1 \choose k} = {n\choose k-2} + 2 {n\choose k-1} + {n\choose k}$. Suppose that $k<m+1 = \frac{n+2}{2}$, then \begin{align*} {2(m+1)\choose k} - {2(m+1)\choose k+1} &= {2m \choose k-2}+{2m\choose k-1}-{2m \choose k}-{2m\choose k+1}\\ &= \left[ {2m \choose k-2} - {2m \choose k} \right] + \left[ {2m\choose k-1} -{2m\choose k+1}\right] \end{align*} The first term is directly strictly negative by induction, the second term is too if $k<m$, if $k=m$ then ${2m\choose m-1}={2m\choose m+1}$ and the second term is $0$. This proves that ${2n+2\choose k}<{2n+2\choose k+1}$. The same can be done when $k>m+1$ to obtain ${2n+2\choose k}<{2n+2\choose k-1}$.

P. Quinton
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