<s>[INST] 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?
 [/INST] 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.
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