Question

Determine the number of proper subsets contained in S if S = x

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`.