How many ways can you split $11$ people into $2$ groups You have 11 students and are creating two groups. At least one student must be in a group. How many different combinations exist. 
My Solution:
We know that no group can be empty, so we put $1$ student in every group. This gives us $9$ renaming students.
Lets say that we only want to add $1$ more student in group $A$, the amount of choices we have is $ 9 \choose 1 $ while all the renaming students are sent to the other group. If we wanted to place an extra two students in the group we would have $ 9\choose2 $ choice, this continues until $ 9 \choose 9 $. 
So the total number of combinations is the sum of the individual options:
$$ {9\choose1}+{9\choose2}+{9\choose3}+\cdots+{9\choose9}$$
I was wondering if my solutions is correct and if there is a better way.
 A: Suppose one of the students is Fred.  Each of the other students must be placed in his group or the other group.  That gives us $2^{10}$ choices.  However, we cannot place all ten of the other students in Fred's group, otherwise the other group would be empty, so there are $2^{10} - 1$ ways to distribute the students to two non-empty groups.
We could correct your solution.  Start with Fred.  We must place either one, two, three, four, five, six, seven, eight, nine, or ten of the other ten students in the other group, giving
$$\sum_{k = 1}^{10} \binom{10}{k} = \sum_{k = 0}^{10} \binom{10}{k} - 1 = 2^{10} - 1$$
ways to distribute the students to two non-empty groups.
A: Your solution is a bit wrong. if you replace the $9$ with $10$ then you will get the right answer, recall from the binomial theorem that
\begin{eqnarray*}
\sum_{n=1}^{10} \binom{10}{n} = 2^{10}-1.
\end{eqnarray*}
So an easier way to this solution would be to say ...
Put the first person into a group & then the other $10$ can be put in either group with the exclusion of putting all $10$ into the first group.
