Simplify sum of combinations with same n, all possible values of k Is there a way to simplify this equation? 
$$\dbinom{8}{1} + \dbinom{8}{2} + \dbinom{8}{3} + \dbinom{8}{4} + \dbinom{8}{5} + \dbinom{8}{6} + \dbinom{8}{7} + \dbinom{8}{8}$$
Or more generally,
$$\sum_{k=1}^{n}\dbinom{n}{k}$$

See 
http://en.wikipedia.org/wiki/Combination#Number_of_k-combinations_for_all_k 
which says 
$$ \sum_{k=0}^{n} \binom{n}{k} = 2^n$$ 
You can prove this using the binomial theorem where $x=y=1$. 
Now, since $\binom{n}{0} = 1$ for any $n$, it follows that 
$$ \sum_{k=1}^{n} \binom{n}{k} = 2^n - 1$$ 
In your case $n=8$, so the answer is $2^8 - 1 = 255$. 
