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How do you prove $n \choose k$ is maximum when $k$ is $\lceil n/2 \rceil$ or $\lfloor n/2 \rfloor$?

This link provides a proof of sorts but it is not satisfying. From what I understand, it focuses on product pairings present in $k! (n-k)!$ term which are of the form $i \times (i-1)$. Since these are minimized when $i=n/2$, we get the result. But what about the reasoning for the rest of the terms?

Rócherz
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curryage
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    Please don't put the question (only) in the title. And, if reasonably possible, make the question self contained by telling what is behind the link. – Magdiragdag Mar 23 '14 at 06:10
  • Have you looked at Pascal's Triangle to see the symmetry that would be the easiest way to see this? The part you are missing is how many factors exist for $k!(n-k)!$ which for small values of $k$ would be near $n$ until you get to the midpoint. – JB King Mar 23 '14 at 06:22
  • @JBKing It is evident "from inspection" of Pascal's Triangle as you point out. I do not understand your second statement. – curryage Mar 23 '14 at 06:27
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    By the "how many factors" I meant the size of that product that when k=0 or k=n, the product is $n!$ that as k gets closer to n/2, the product gets smaller as one is taking out the biggest factor and adding a new smaller factor usually. Consider $n!$ versus $(n-1)!$ verses $(n-2)!*2$ etc. – JB King Mar 23 '14 at 06:33
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    See here: http://math.stackexchange.com/questions/88144/how-can-i-prove-that-the-binomial-coefficient-n-choose-k-is-monotonically-n – Martin Sleziak Mar 23 '14 at 09:29
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    Link in question is broken – Colm Bhandal May 21 '20 at 19:14

6 Answers6

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HINT:

As $\displaystyle \binom nk>0$ for $0\le k\le n$ where $n>0,k$ are integers

Check for $k$ such that $$\frac{\binom n{k+1}}{\binom nk}=\frac{n-k}{k+1}>=<1$$

lab bhattacharjee
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    Is $>=<$ the long-lost "bow-tie" operator? – Mario Carneiro Mar 23 '14 at 07:33
  • @lab bhattacharjee This lets us know the conditions under which ratio is >=< 1, i.e when $k <=> \frac{(n-1)}{2}$. But how do we go from here to proving a fact about $n \choose k$ in isolation ? – curryage Mar 23 '14 at 07:39
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    @curryage Note that "${n\choose k}$ is maximum at this $k$" is not a fact about ${n\choose k}$ in isolation, but rather by comparison to all other $k'$ in the range of the function. See also my answer. – Mario Carneiro Mar 23 '14 at 08:01
  • I know I'm over 7 years late, but I can't find anywhere what "$>=<$" means. – Just_A_User Oct 17 '21 at 01:04
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    @JamesA It just means check whether it greater or lesser than (or equal to) 1, which would tell you when the sequence is increasing or decreasing – Anshuman Agrawal Feb 15 '22 at 08:15
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I have done this proof in Metamath before; it may help to see the whole thing laid out.

The proof follows from the fact that the binomial coefficient is monotone in the second argument, i.e. ${n\choose k'}\le{n\choose k''}$ when $0\le k'\le k''\le\lceil\frac n2\rceil$, which can be proven by induction. Given this, you just set $k''=\lceil\frac n2\rceil$ and $k'=k$ or $k'=n-k$ depending on whether $k\le\frac n2$, and you get ${n\choose k}={n\choose n-k}\le{n\choose \lceil n/2\rceil}={n\choose \lfloor n/2\rfloor}$ (where the equalities are deduced by symmetry of the binomial coefficient under $k\mapsto n-k$).

To prove monotonicity, we prove ${n\choose k-1}\le{n\choose k}$ for $1\le k\le\lceil\frac n2\rceil$, and thus ${n\choose k}\le{n\choose k+1}\le\dots\le{n\choose l}$ for any $k\le l$ in the range. Now we have:

$${n\choose k-1}=\frac{n!}{(k-1)!(n-k+1)!}=\frac{n!}{k!(n-k)!}\frac{k}{n-k+1}={n\choose k}\frac{k}{n-k+1},$$

so ${n\choose k-1}\le{n\choose k}$ iff $\frac{k}{n-k+1}\le 1$. But that is equivalent to $$k\le n-k+1\iff 2k\le n+1\iff k\le \frac{n+1}2\iff k\le \left\lceil\frac{n}2\right\rceil,$$

and we are done.

Mario Carneiro
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  • Thank you very much for the great answer! I am just very curious as a beginner and will be glad if you answer this. How do you know how to start, where to approach from in order to prove this? Do you have some intuition before starting, or you just write down known things and prove after several attempts? – Turkhan Badalov Aug 26 '17 at 21:02
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    @TurkhanBadalov This proof is very much driven by a bit of knowledge of Pascal's triangle. If you look at it you will notice that it is symmetric, and in each row the numbers go up until you reach the center; from this you can conjecture the statement. To prove it, you want a way to relate nearby binomial coefficients, and the fact that it is a product of factorials means that there is a nice formula for adding one in any direction, and Wikipedia will supply ${n\choose k}=\frac{n+1-k}{k}{n\choose k-1}$. When the fraction is greater than 1, the numbers are increasing, else they are decreasing. – Mario Carneiro Aug 27 '17 at 01:35
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Here's a combinatorial proof, by counting pairs of subsets contained in one another:

Any $k$-element subset is contained in $n-k$ different $(k+1)$-element subsets, whereas each $(k+1)$-element subset contains exactly $(k+1)$ different $k$-element subsets. So provided that $n-k > k+1$ we have that $\binom nk < \binom n{k+1}$, and an inequality the other direction in the other case. The fact that the maximum is in the middle follows immediately.

Nate
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The second identity here gives $$\binom{n}{k}-\binom{n}{k-1}=\frac{n+1-2k}{n+1} \binom{n+1}{k}. $$ Thus the sequence of binomial coefficients $\binom{n}{k}$, with $n$ fixed, is increasing as long as $2k \leq n+1$ or $$k \leq \left\lfloor \frac{n+1}{2} \right\rfloor, $$ and is decreasing for all values of $k$ satisfying $2k \geq n+1$ or $$k \geq \left\lceil \frac{n+1}{2} \right\rceil.$$

user1337
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There is a general trick for functions with symmetry which can be used here if you allow for non-integer continuation of the factorial function (which one should I think). Let us denote:

$\binom{n}{k} = F_{n}(k)$.

The binomial identity give us:

$ F_{n}(k) = F_{n}(n-k)$.

Taking the derivative of both sides gives:

$F'_{n}(k)=-F'_{n}(n-k)$.

Now assuming that only one extrema exists at $k = \bar{k}$ then $F'(\bar{k}) = 0$. We then have:

$F'(n-\bar{k}) = 0 \rightarrow \bar{k} = (n-\bar{k})$

Giving, $\bar{k} = n/2$. Translating to integers this gives the floor and ceil of (n/2) as the solutions.

Of course I had to assume there was only extrema of the binomial coefficient. One can see this by graphing the binomial coefficient but here is a sketchy outline of how it can be done with these functions.

From the requirements, $F_{n}(0) = F_{n} (n) = 1$ and $F_{n}(k)>1$, we know that the number of extrema must be odd. From the binomial identity,

$F_{n}(k+1) = \frac{n-k}{k+1} F_{n}(k)$,

one can show that the derivative an integer step to the left/right of the extremal point is positive/negative. This shows that the only extremal points allowed are concave. Since one can't have more than one extrema without having convex extrema, we conclude that there is only one extrema.

jman
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We can write $$\binom{n}k =\frac{n!}{(n-k)! k!}=\frac{\Gamma(n+1)}{\Gamma(n-k+1)\Gamma(k+1)},$$ where $\Gamma$ is the Gamma-function.

I will prove something slightly different, but related to the question, namely that, in the above sense, we have $$\binom{n}{\frac n2}\ge \binom{n}k$$ for all $k\in[0,n]$. and $n\in[1,\infty[$.

For this note that we only need to prove $$\Gamma(n-k+1)\Gamma(k+1)\ge\Gamma(n/2+1)^2.$$ After taking logarithms, this is just $$\frac12\left(\ln(\Gamma(n-k+1))+\ln(\Gamma(k+1))\right)\ge\ln(\Gamma(n/2+1)),$$ which follows immediately from log-convexity of $\Gamma$.

Maximilian Janisch
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