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There is the set of unborn children who were [[abortion|aborted]], about which striking conclusions can be drawn.  Given the large and diverse number of elements of this set, it would likely include many who could surpass existing athletic and intellectual achievements.  Indeed, many of the world records and [[Nobel Prize]] achievements recognized today would have been outdone by members of this set.

There is the set of unborn children who were [[abortion|aborted]], about which striking conclusions can be drawn.  Given the large and diverse number of elements of this set, it would likely include many who could surpass existing athletic and intellectual achievements.  Indeed, many of the world records and [[Nobel Prize]] achievements recognized today would have been outdone by members of this set.

Another striking example is the how traditional marriage provides a greater set than otherwise:  the union of <math>A = \{a, b, c, d\}</math> and <math>B = \{a, b, c, e\}</math> is merely <math>\{a, b, c, d, e\}</math>, while the union of a man, <math>M = \{a, b, c, d\}</math> and a woman <math>W = \{e, f, g, h\}</math> is <math>\{a, b, c, d, e, f, g, h\}</math>, which is a broader and more diverse set.  

Another striking example is the how traditional marriage provides a greater set than otherwise:  the union of <math>A = \{a, b, c, d\}\,</math> and <math>B = \{a, b, c, e\}\,</math> is merely <math>\{a, b, c, d, e\}\,</math>, while the union of a man, <math>M = \{a, b, c, d\}\,</math> and a woman <math>W = \{e, f, g, h\}\,</math>, is <math>\{a, b, c, d, e, f, g, h\}\,</math>, which is a broader and more diverse set.  

Define "A" as the set of all false assertions, and "B" as the set of all the Bible passages. Since there are no [[Counterexamples to the Bible]], the intersection between set A and set B is the [[empty set]].

Define "A" as the set of all false assertions, and "B" as the set of all the Bible passages. Since there are no [[Counterexamples to the Bible]], the intersection between set A and set B is the [[empty set]].

==More Applications -- Combinatorics==

==More Applications -- Combinatorics==

Sets, and their cardinalities, are used in some very important aspects of combinatorics, that is, the study of combinations.  Let <math>X = \{a,b,c\}</math> and <math>Y = \{p,q,r\}</math>.  The cardinalities of both are <math>\|X\| = \|Y\| = 3</math>.  Let <math>S = \{\{x,y\}\ |\ x \in X \and y \in Y\}</math>.  The vertical bar means "such that", and the epsilon-like symbol means "is an element of".  So this means that S is the collection of sets<ref>Mathematicians don't like to use the phrase "set of sets", since this can lead to some unpleasant paradoxes.  So they say "collection of sets."  A "collection" is really just a set.</ref> which are pairs x and y such that x is in X and y is in Y.  We can enumerate this.  <math>S = \{\{a,p\}, \{a,q\}, \{a,r\}, \{b,p\}, \{b,q\}, \{b,r\}, \{c,p\}, \{c,q\}, \{c,r\}\}</math>.  Its cardinality is <math>\|S\| = 9</math>.

Sets, and their cardinalities, are used in some very important aspects of combinatorics, that is, the study of combinations.  Let

:<math>X = \{a,b,c\}\,</math> and <math>Y = \{p,q,r\}\,</math>.

Now try a harder example, based on the same two sets.  Let <math>G = \{F\ |\ F \subset (X \cup Y) \and \|F\|=2\}</math>.  Now <math>X \cup Y = \{a,b,c,p,q,r\}</math>, so <math>\|X \cup Y\| = 6</math>.  The problem of finding the number of subsets of a set that has some particular cardinality is the ''combination choosing'' problem, and shows up in many areas of probability and statistics.  In this case, it's the problem of choosing 2 elements out of a set of 6 elements.  The number of such choices is often written <math>{}_6C_2</math> or <math>6 \choose 2</math>, and is often pronounced "six choose two".  These numbers are also called the [[binomial coefficients]] because they appear as coefficients in the expansion of <math>(x+y)^n</math>.  In this case, <math>6 \choose 2</math> is the coefficient of <math>x^2y^4</math> in <math>(x+y)^6</math>.

The cardinalities of both are <math>\|X\| = \|Y\| = 3</math>.

There is a simple formula for <math>n \choose k</math>.  It is <math>{n \choose k} = \frac{n!}{n! (n-k)!}</math>. In the case of the set <math>G = \{F\ |\ F \subset \{a,b,c,p,q,r\} \and \|F\|=2\}</math>, the set is <math>G = \{\{a,b\}, \{a,c\}, \{a,p\}, \{a,q\}, \{a,r\}, \{b,c\}, \{b,p\}, \{b,q\}, \{b,r\}, \{c,p\}, \{c,q\}, \{c,r\}, \{p,q\}, \{p,r\}, \{q,r\}\}</math>. Its cardinality is 15.

 

 

 

There is a simple formula for <math>n \choose k</math>.  It is

 

:<math>{n \choose k} = \frac{n!}{n! (n-k)!}</math>.

 

In the case of the set <math>G = \{F\ |\ F \subset \{a,b,c,p,q,r\} \and \|F\|=2\}</math>, the set is

:<math>G = \{\{a,b\}, \{a,c\}, \{a,p\}, \{a,q\}, \{a,r\}, \{b,c\}, \{b,p\}, \{b,q\}, \{b,r\}, \{c,p\}, \{c,q\}, \{c,r\}, \{p,q\}, \{p,r\}, \{q,r\}\}\,</math>.

Its cardinality is 15.

We can check that <math>{6 \choose 2} = \frac{6!}{2! (6-2)!} = \frac{720}{2 \cdot 24} = 15</math>.

We can check that <math>{6 \choose 2} = \frac{6!}{2! (6-2)!} = \frac{720}{2 \cdot 24} = 15</math>.