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Understanding Inclusion/Exvlusion with Venn Diagrams

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We want to know the total of x 1 +x 2 +x 3. Problem is there is some overlap. Some of x 1 is in x 2, some of x 1 is in x 2, some of x 2 is in x 3,and finally, there are some that are in all 3. We know that x 1 has X amount, x 2 has Y amount, and x 3 has Z mount. The question is, “How many in total are there?” We’ll have to use the process of Inclusion and Exclusion.

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Lets define some of these areas.

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First we’ll start off by adding in the total of what ever x1 is. Formula so far: N(x1 V x2 V x3) = N(x1)

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Next we’ll add in x2. This will cause some overlap between x1 and x2. This is ok because we’ll fix this later. Formula so far: N(x1 V x2 V x3) = N(x1) + N(x2)

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Next we’ll add in x3. This will cause some overlap between [x1 & x3] & [x2 & x3] & and [x1 & x2 & x3]. Formula so far: N(x1 V x2 V x3) = N(x1) + N(x2) + N(x3)

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Now we’ll fix the overlap between [x1 & x2]. When there was overlap & this could equate to [x1 & x2]+[x1 & x2] or 2([x1 & x2]). When we do a subtraction of the middle section, we’re left with only one piece of [x1 & x2]. Notice how it’s blue in the center between [x1 & x2]? That’s because we took out the red portion of it. Also, the section of [x1 & x2 & x3] is now teal because the red was taken out from that portion too Formula so far: N(x1 V x2 V x3) = N(x1) + N(x2) + N(x3) – N(x1 & x2)

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Now we’ll fix the overlap between [x2 & x3]. Similar to the removal of [x1 & x2], this will leave only the green section. The section of [x1 & x2 & x3] is now green because the blue was taken out from that portion. Formula so far: N(x1 V x2 V x3) = N(x1) + N(x2) + N(x3) – N(x1 & x2) – N(x2 & x3)

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Now we’ll fix the overlap between [x1 & x3]. Similar to the removal of [x1 & x2], this will leave only the red section. The section of [x1 & x2 & x3] is now empty. That’s because in the process of removing the common ground, we removed the center as well. Luckily, there’s a way to fix it! Formula so far: N(x1 V x2 V x3) = N(x1) + N(x2) + N(x3) – N(x1 & x2) – N(x1 & x3) – N(x2 & x3)

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Thankfully, fixing the gap in that hole isn’t as difficult as you’d think. It’s just a matter of adding in [x1 & x2 & x3]! That’s it! Now you have the total of all the x1, x2, and x3. Formula so far: N(x1 V x2 V x3) = N(x1) + N(x2) + N(x3) – N(x1 & x2) – N(x1 & x3) – N(x2 & x3) + N(x1 & x2 & x3)

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