<p>Well if I'm not mistaken, the set the other poster listed would not contain D multiple times.</p>
<p>A set is a group of items. An example of two sets we'll define as such
Set A = {5,6,9,22}
Set B = {6,7,8,9,10}</p>
<p>Now, the union of two sets, is one set containing each item with NO repeats. (capitalization for accentuation),
Therefore the A∪B would be {5,6,7,8,9,10,22}.</p>
<p>This can be illustrated by the diagram below:
<a href="http://upload.wikimedia.org/wikipedia/commons/2/2f/Venn_A_union_B.png%5B/url%5D">http://upload.wikimedia.org/wikipedia/commons/2/2f/Venn_A_union_B.png</a></p>
<p>Now -just to go a little further- I will go into intersections.</p>
<p>Intersections are actually the opposite of a union. They are all the items in BOTH sets (The items that they have in common) . The A∩B of our example sets would be: {6,9}</p>
<p>Now, on the SAT they normally won't mention sets in a problem at all if it can be solved using them. Here is a quick and dirty example problem that most questions will be similar to:</p>
<p>In my math class over the course of the year, 12 students have gotten A's on my tests, 16 have gotten B's, and 6 have gotten both A's and B's on my tests. No one got a lower grade then a B at any point. How many students do I have in my class?</p>
<p>Tackling this problem:
First lets reiterate what the problem tells us. Out of all the students in the class, 12 have gotten an A on a test at some point, 16 have gotten a B at some point, and 6 have gotten both grades at some point. </p>
<p>Normally with the SATs, you can solve these problems with this formula:
Total=(number of items in set A)+(number of items in set B)-(number of items in set C [your items in common set])</p>
<p>Using this formula we can simplify this problem into this:
Total=12+16-6</p>
<p>This means that there is 22 students in the class.</p>
<p>But, just how does this work? I haven't seen a traditional proof, but this would be my attempt to prove it:</p>
<p>This would be the set of students that have gotten an A-> {Jill, Ben, Sue .... Fred}
This would be the set of students that have gotten a B -> {Ben, Sue, .... Alex}
This would be the set of students that have gotten both an A and a B -> {Ben, Sue ....}
(sets are NOT limited to numbers, but in math usually contain numbers)</p>
<p>Now, we have a set of students who have gotten A's, B's, and both. Well, our set containing A's contains all the students who have gotten A's only, and both A's and B's. Set B contains all the students who got only B's, and both A's and B's. Therefore we are looking at something like this:</p>
<h1>of kids with only A's + # of kids with both = # of entries in set A</h1>
<h1>of kids with only B's + # of kids with both = # of entries in set B</h1>
<p>Now, if we were to add these together based upon only the number of entries each set has, we'd have something like this:</p>
<p>(# of kids with only A's + # of kids with only B's + 2(# of kids with A's and B's) = # of entries in set A + number of entries in set B</p>
<p>So, in order to find our answer, we need to get rid of the duplicate ammount of kids who have both A's and B's. That's where we subtract the ammount of kid's who only have A's and B's, given with the number of entries in set C.</p>
<p>So we wind up with the formula:</p>
<h1>of kids with only A's + # of kids with only B's + # of kids with both A's and B's = # of entries in set A + number of entries in set B - number of entries in set C</h1>
<p>Hopefully that helps.</p>
