Prove: the biggest that N choose k gets is at k = N/2 for N even

[AxiomOfChoice]

Is there a simple way to prove this? I tried to use induction, but you get down to some horrible fraction (letting $N = 2m$ for some $m$) in the inductive step:

$$\begin{pmatrix} 2(m+1) \\ k \end{pmatrix} = \frac{(2m)!}{k!(2m - k)!} \cdot \frac{(2m+2)(2m+1)}{(2m+2-k)(2m+1-k)}$$

The inductive hypothesis is to assume the thing on the left is biggest for $k = m$, but the second fraction gets bigger as you make $k$ bigger. So...what to do! Any comments? Thanks!

 

[R136a1]

Try to prove this: If $k<k^\prime\leq N/2$, then

$$\binom{N}{k} < \binom{N}{k^\prime}$$

And if $N/2\leq k^\prime<k$, then

$$\binom{N}{k} < \binom{N}{k^\prime}$$

To prove these two statements, it suffices to look at $k$ and $k+1$.

 

[Erland]

Note that it suffices to show that k!(2m-k)! attains its minimum at k=m (0≤k≤2m), and this is equivalent to showing that
(m-r)!(m+r)! attains its minimum at r=0 (0≤r≤m).

 

[BTP]

Is there a simple way to prove this?

let n=even, k<=n/2
$$\begin{pmatrix} n \\ k \end{pmatrix} =\begin{pmatrix} n \\ k-1 \end{pmatrix} *{ (n-k+1)/(k)}$$

then (n-k+1)/k > (n/2 +1)/k >1

QED

...the symmetry between k and n-k finishes the proof.

 

[blue_raver22]

There is indeed a simple way to prove this. First, let's rewrite the expression for N choose k in terms of factorials:

\begin{pmatrix} N \\ k \end{pmatrix} = \frac{N!}{k!(N-k)!}

Now, let's consider the ratio of N choose k and N choose k+1:

\frac{\begin{pmatrix} N \\ k \end{pmatrix}}{\begin{pmatrix} N \\ k+1 \end{pmatrix}} = \frac{\frac{N!}{k!(N-k)!}}{\frac{N!}{(k+1)!(N-k-1)!}} = \frac{(k+1)(N-k)}{(N-k-1)(k+1)} = \frac{k+1}{N-k-1}

We can see that this ratio is always greater than 1 when k < N/2, and less than 1 when k > N/2. This means that the value of N choose k increases until k = N/2, and then decreases. Therefore, the maximum value of N choose k occurs at k = N/2 for N even.

As for the inductive step, we can simply use the fact that the ratio of N choose k and N choose k+1 is always greater than 1 for k < N/2, and less than 1 for k > N/2. This means that the value of N choose k+1 is always smaller than N choose k when k < N/2, and larger when k > N/2. Therefore, the maximum value of N choose k+1 must occur at k = (N+1)/2, and since N is even, this reduces to k = N/2.

In conclusion, the maximum value of N choose k occurs at k = N/2 for N even, and this can be easily proven by considering the ratio of N choose k and N choose k+1.