Question

Determine the number of proper subsets contained in S if S = {x|x is an even, negative integer and x > -27}

Answer 1

First, lets determine $|S|$ (the number of elements in $S$).

We can see that the negative even integers that are between -27 and 0 are: -24, -22, -20, -18, -16, -14, -12, -10, -8, -6, -4, and -2.
This means $|S|=13$.

A proper subset of $S$ is can't be $S$ itself, this means the a proper subset of $S$ can't have 13 elements.

Given that re-arranging the elements of any set doesn't yield a different set, we use combinations (binomial coefficients) to get the number of subsets in a particular set.

The number of subsets of $S$ with 1 element are given by $(^{13}_{1})$ (which means 13 combine 1), in the same way the number of subsets with 2 elements is given by $(^{13}_{2})$ and so forth until reaching the number of subsets with 12 elements.

The previous statement can be written as:
$\Sigma^{12}_{k=0}(^{13}_{k})$

In the previous summation we also included the empty subset, which is always a proper subset of $S$.

So $\Sigma^{12}_{k=0}(^{13}_{k})$ is an acceptable answer, but we can simplify it. 
The following statement is true (it can be easily proven):
$\Sigma^{n}_{k=0}(^{n}_{k})=2^n$

From our previous discussion, this means that the number of subsets of any arbitrary set $S$ is equal to $2^{|S|}$.
This means that the number of proper subsets (which is the number of subset minus the set itself, for finite subsets) is equal to $2^{|S|}-1$.

So, the answer is:
$\Sigma^{12}_{k=0}(^{13}_{k})=2^{13}-1=8192-1=8191$

There are 8191 proper subsets of $S$.