Combinatorics: Amount of options for majority Say we have $2n$ people. Then the amount of options to form a majority (e.g. in a commission) are
$ \binom{2n}{n+1} + \binom{2n}{n+2} + \cdots + \binom{2n}{2n}$
I want to prove, that this is equal to $\frac{1}{2} \left[ 2^{2n} - \binom{2n}{n} \right]$
But I'm stuck. I have the this formula: $ \binom{n}{0} + \binom{n}{1} + \cdots + \binom{n}{n} = 2^n $
So I can say $ \binom{2n}{n+1} + \binom{2n}{n+2} + \cdots + \binom{2n}{n} = 2^{2n} - \left[ \binom{2n}{n} + \binom{2n}{n-1} + \cdots + \binom{2n}{n-n} \right] = 2^{2n} - \left[ \binom{2n}{0} + \binom{2n}{1} + \cdots + \binom{2n}{n} \right] $
But I can't see where to go from there, even when writing the binomial coefficients as $ \frac{n!}{k!(n-k)!} $
Can anyone help me with it?
 A: By symmetry $\binom{2n}{k}=\binom{2n}{2n-k}$ for $k=0,\ldots,n-1.$
So the sum, call is $S$, you have is also equal to
$$
\binom{2n}{0}+\binom{2n}{1}+\cdots+\binom{2n}{n-1}.
$$
Now $2S+\binom{2n}{n}=2^{2n},$ does it.
A: *

*There are $2^{2n}$ ways of selecting a subset to vote "Yes" and the rest "No".


*Of these, ${2n \choose n}$ are the number of ways to have equal "Yes" and "No" votes,


*leaving  $2^{2n}-{2n \choose n}$ possibilities with a decisive result,


*half of which have a majority voting "Yes", getting your result
(the other half of the possibilities would have a majority voting "No"; a particular subset of individuals with the majority would be double counted, once when voting "Yes" against the rest and once when voting "No" against the rest).
A: By the property of binomial coefficients $$\binom{n}{r}=\binom{n}{n-r}$$
$\implies \binom{2n}{n+1}=\binom{2n}{n-1}$
$\implies\binom{2n}{n+2}=\binom{2n}{n-2}$ and so on

Let $$A=\binom{2n}{n+1}+\cdots+\binom{2n}{2n}$$ Also
$$A=\binom{2n}{0}+\cdots+\binom{2n}{n-1}$$
So $$2A=2^{2n}-\binom{2n}{n}$$
