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

User Sereja
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1 Answer

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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.
User Kevin Lamb
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