Understanding Inclusion/Exvlusion with Venn Diagrams

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.

Lets define some of these areas.

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)

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)

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)

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)

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)

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)

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)