Start by explicitly writing out the members of each set:

Part A

To take the intersection, means to include only the candies on both podium and , so:

Because is the only candy that is included in both of these sets.

Part B

Start with the elements in and then remove the elements in , so:

Only occurred in both of the sets.

Part C

Start with the elements in this time, then remove the elements from , this is different to , so:

Again, only occurs in both.

Part D

To take the union of the sets, just include all of the candies that show up on both podiums and , so:

These are the only two that show up on both podiums.

Unions

Unions include every element in either set or , so it contains all elements without duplicates, remember that order doesn’t matter:

Intersections

Intersections contain all elements that show up in both and :

Part a

The children that like ice cream are all of the children in set :

Part b

The children that don’t like gum can be found with :

Part C

The intersection between these can be shown on the diagram, so:

Part D

As we can see, this resulting set includes all of the children that like ice cream but do not like gum, so the resulting set is given by:

This theorem applies in general to any two sets.

First, we need to realize that b appears in both sets, so it has to be drawn in the intersection of and . The rest of the elements are as shown:

Part A

The set contains all elements in or :

Part B

The set contains all elements in and

Part C

Here, E is a number, we just want the cardinality of , so:

Part A

Here we want people in set or set . So these people must either like chocolate, like candy or they can like both.

Part B

Here we want people in both set and , so they need to like both chocolates and sweets.

Part C

In order to define , we need to define a universal set first. So we can just let be a set that contains all people.

Once we do that, is just the set of people who don’t like chocolate.

Do those people even exist?

We will show two different ways to solve this problem. So we will explore both. However in either case, we need to put everybody into a set, so we define:

• U: The universal set of all of the “volunteers” 😉

• S: The set of people who were successfully shrunk

• P: The set of people who developed purple tongues

We should then show this information in a Venn diagram:

This let’s us say that:

First Method (Direct)

Since 17 people shrunk, and only 10 got purple spots, then 7 did not get purple spots, therefore:

So we can just break up these sets into three disjoint parts:

This means that:

This means that people got purple spots without shrinking, so:

So that’s 7 people with purple spots on their tongue that didn’t shrink.

Second Method (Inclusion Exclusion)

We also have all the information necessary to solve this with the inclusion and exclusion principle. First we solve for the total number of people with purple spots , so:

So that means that the missing number is 17, since , so:

Now if we know that 10 of these people also shrunk, since , then we have:

Then the missing number is just 7 again, so:

Which is the same conclusion we came to last time.

In this case, Oompa starts with a universal set of 25 materials, so we can say that:

We should also define the sets:

• : The set of materials that are strong enough

• : The set of materials that are colorful enough

Now since 6 materials have been outright rejected, we can see that 19 materials are left in the possible pool of materials that belong in either or , so we can say that:

We can draw this with a Venn diagram:

So we just need to find out how many materials go into each set.

Inclusion and Exlusion

So we can use the inclusion and exclusion principle to solve the problem directly now:

We have all of these numbers, so we just put them directly in:

So there are 6 possible materials to choose from.

To solve this problem, we need to start by assigning every possibility a set. So let’s denote these with:

• The set of joggers can be
  

• The set of swimmers can be
  

• The set of footballers can be
  

We can draw this in a Venn Diagram where these sets may have some overlap:

To solve the problem, just use the principle of inclusion and exclusion. This will give:

We know that 15 were inactive, so , which means that 70 people surveyed did some form of physical activity. This means that:

So the missing value is , which is 12, so:

So there are 12 people who partake in all three types of physical activity.

Part A

We just need to include everything that is not in the set intersection, so:

Part B

We can show both of these on two separate Venn Diagrams

Part C

Showing the set union of these sets includes all members of both sets.

Part D

We can see that the two sets are identical to each-other. The only elements that are not included are the elements in the set union.

Part E

Then Oompa is referring to the set

This is clear because contains all chocolates rejected by group A and contains the chocolates rejected by group B, and Oompa said that they are talking about the intersection with the word or.

To simplify the statement, just apply De Morgan’s law, so we get:

Now is the group of chocolates that both group A and B liked. So Oompa’s statement just becomes:

There were some chocolates that were rejected by both group A and B

You should try picking elements in any of the regions carved out by the Venn diagram to convince yourself that these sentences mean the exact same thing.

Both rules can be explained by a simple observation, that if (X is a subset of Y), then we can say:

Subset observations

To justify this, we can say that:

• The first observation is true because if something is in X or Y, it is always in Y since Y surrounds X.

• The second is true because if something is in X and Y, then it must always be in X, because X is inside of Y.

First Law

The first law is true because is a subset of , so therefore the union will include the larger set, so the result is just A, hence:

Second Law

The second law is true because is a subset of , so their union will equal the smaller set. Since A is inside of , the result is just A again, so:

Part A

We start by drawing out each step on the left and the right hand side. Notice that the final set in both cases ends up being equal:

Part B

If this statement is true. Then we let:

• The set of friends be
  

• People who play sports belong to
  

• People who play video games belong to
  

So therefore:

You can read the right hand side as:

“I want to go out with” My friends up the road (F), who either like playing sports (S) or playing video games (V)