How to demonstrate that (MAX Z = (100 choose X)), is X = 50

Question

How to demonstrate that $\max(Z)$ where $\displaystyle Z=\left\{{100 \choose X}\right\}_{X\in[[0..100]]}$, has X taking the value of 50? Intuitively makes sense but not sure how to prove that? Proving with English is fine.

Answer 1

Well, consider that:

$$\dfrac{100!}{49!~ 51!}=\dfrac{100!}{50!~ 50!}\dfrac{50}{51}$$

So clearly $\dbinom{100}{49}< \dbinom{100}{50} > \dbinom{100}{51}$

Show $\dbinom{100}{k}<\dbinom{100}{50}$ holds for any $k$ lesser or greater than $50$.

Answer 2

Generalized version: for n even,

$(\frac{n}2 + 1)(\frac{n}2 - 1)! > (\frac{n}2)(\frac{n}2 - 1)!$

$(\frac{n}2 + 1)!(\frac{n}2 - 1)! > (\frac{n}2)!(\frac{n}2)!$

$\frac{1}{(\frac{n}2 + 1)!(\frac{n}2 - 1)!} < \frac{1}{(\frac{n}2)!(\frac{n}2)!}$

$\frac{n!}{(\frac{n}2 + 1)!(\frac{n}2 - 1)!} < \frac{n!}{(\frac{n}2)!(\frac{n}2)!}$

${n \choose {\frac{n}{2}}} > {n \choose {\frac{n}{2} - 1}}$.

Also, ${n \choose {\frac{n}{2} - 1}} = {n \choose {\frac{n}{2} + 1}}$, gives you the same inequality in the other direction. Then proceed by induction. Or just replace the "-1" in the first line with "-k" for k $\geq$ 1 and prove it that way

Answer 3

$$\frac{n\choose k}{n\choose k-1}=\frac{n-k+1}{k}$$

If $k\le n/2$, this is $>1$. And if $k\ge n/2+1$, it's $<1$.

One case is left: if $n$ is odd and $k=\lceil n/2 \rceil$, then $\frac{n-k+1}{k}=1$.