Combinations in sets

Question

Say a set $X=\{A,B,C,D\}$

How many proper subsets does it have? And how do you determine this? I thought that it was 14 due to adding up combinations of:

$4C_1 + 4C_2 + 4C_3$ so $4+6+4=14$ possibilities

However, it turns out to be $15$ in the answer of my textbook.

What is the $15^{th}$ possibility? and also how do you determine the pattern for these kinds of sets? (for example set $Y=\{A,B,C,D,E,F\}$ with $6$ values and set $Z=\{A,B,C\}$ with $3$ values)

How can $A$ be in also $A$? Names are problematic. – May 26 '17 at 09:43 — , May 26 '17 at 09:43
Oh, Some error. Thanks for mentioning :) – nabu1227 May 26 '17 at 09:46 — nabu1227, May 26 '17 at 09:46

CY Aries · Accepted Answer · 2017-05-26T09:53:27.843

3

You missed the empty set, which is also a proper subset.

To form a subset, each of $A$, $B$, $C$, $D$ can be either chosen or not chosen as an element. So you have 2 choices for each of $A$, $B$, $C$, $D$. There are totally $2\times 2\times 2\times 2=2^4$ choices. Excluding $X$ itself, there are $2^4-1$ proper subsets.

edited May 26 '17 at 09:53

answered May 26 '17 at 09:38

CY Aries

23,393

Why the downvote??? – Parcly Taxel May 26 '17 at 09:40
Confused me too. – Arby May 26 '17 at 09:40
Hold on, if a subset isn't a proper subset, does it still include the empty set. If no, why? – nabu1227 May 26 '17 at 09:41
http://mathworld.wolfram.com/ProperSubset.html. The empty set is a proper subset, $X$ itself is not. That's why we only have $2^4-1$ proper subsets, as Mickey mentioned. – CY Aries May 26 '17 at 09:45

Combinations in sets

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