Show that $2^n ≤$ $n\choose n/2$ is contradictory

Question

I am writing a proof and I am having issues with this part of the proof. I want to show that $2^n ≤$ $n\choose n/2$ is contradictory.

Does this answer your question? Prove that $2^n < \binom{2n}{n} < 2^{2n}$ — player3236, Jan 29 '21 at 11:21

score 1 · Answer 1 · answered Jan 29 '21 at 11:20

1

Hint: $$ (1+1)^{2m}=\sum_{k=0}^{2m}{2m\choose k}1^k1^{2m-k}$$

answered Jan 29 '21 at 11:20

Hagen von Eitzen

382,479

score 1 · Accepted Answer · edited May 28 '21 at 19:07

1

This is clear by a very simple combinatorial argument: Say $S=\{1,2\dots,n\}$. Then $2^n$ is the total number of subsets of $S$, while $C(n,n/2)$ is...

edited May 28 '21 at 19:07

quid

42,835

answered Jan 29 '21 at 11:29

David C. Ullrich

92,839

Is C(n,n/2) is the number of elements chosen from set S? – Desmond Jan 29 '21 at 11:41
Ah okay i get it! The combination of elements chosen from set S will always be a subset of the total number of subsets possible, hence it is not possible for the combination of elements chosen from set S to be greater than the total number of subsets possible – Desmond Jan 29 '21 at 12:01

Show that $2^n ≤$ $n\choose n/2$ is contradictory

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