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

  • #1

Main Question or Discussion Point

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

[tex]
\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)}
[/tex]

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

Answers and Replies

  • #2
341
51
Try to prove this: If ##k<k^\prime\leq N/2##, then

[tex]\binom{N}{k} < \binom{N}{k^\prime}[/tex]

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

[tex]\binom{N}{k} < \binom{N}{k^\prime}[/tex]

To prove these two statements, it suffices to look at ##k## and ##k+1##.
 
  • #3
Erland
Science Advisor
738
136
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).
 
  • #4
BTP
9
1
Is there a simple way to prove this?
let n=even, k<=n/2
[tex]
\begin{pmatrix} n \\ k \end{pmatrix} =\begin{pmatrix} n \\ k-1 \end{pmatrix} *{ (n-k+1)/(k)}
[/tex]

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

QED

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

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