Question

Of the members of three athletic teams in a certain school, 21 are on the basketball team, 26 on hockey teams, and 29 on the football team. 14 play hockey and basketball, 15 play hockey and football, 12 play football and basketball, and 8 play all three games. How many members are there in all?

Answer 1

Hint: We will categorize the students into sets according to the game they play. Then we will be able to describe the students playing two games as the intersection of two sets and the students playing all three games will lie in the intersection of all three sets. Using the cardinality of these sets, we will be able to find the total number of members.

Complete step by step solution:

Let $\text{A}$ be the set of members who are on the basketball team. So, $|\text{A }\\!\\!|\\!\\!\text{ = 21}$. Let us denote the set of members who are on the hockey team by $\text{B}$. So, the cardinality of $\text{B}$ is 26. Let $\text{C}$ denote the set of members who are on the football team. Therefore, we have $|\text{C }\\!\\!|\\!\\!\text{ = 29}$.

Next, we know that 14 members play both hockey and basketball. These members belong to both the set $\text{A}$ and set $\text{B}$. This means, $|\text{A}\cap \text{B }\\!\\!|\\!\\!\text{ = 14}$. Similarly, there are 15 members who play hockey and football. So, these members are in the intersection of set $\text{B}$ and set $\text{C}$. Hence, $|\text{B}\cap \text{C}|\text{ = 15}$. In the same way, we have 12 players who play football and basketball. This implies that these players are in the intersection of set $\text{A}$ and set $\text{C}$. So, we have $|\text{A}\cap \text{C}|\text{ = 12}$.

Now, we know that there are 8 members who play all three games, which means that these members are in the intersection of all three sets. Therefore, $|\text{A}\cap \text{B}\cap \text{C}|\text{ = 8}$.

We have to count the total number of members. To do this, we will start by adding the number of members in set $\text{A}$, set $\text{B}$ and set $\text{C}$. So, we have $|\text{A }\\!\\!|\\!\\!\text{ + }\\!\\!|\\!\\!\text{ B }\\!\\!|\\!\\!\text{ + }\\!\\!|\\!\\!\text{ C }\\!\\!|\\!\\!\text{ = 21+26+29 = 76}$. But, in this case, we have counted the members who play two games two times, once for one game and the second time for the other game. So, to count these members only once, we will subtract the total number of members who are part of two teams.

The total number of members who are part of two teams is given by,

$|\text{A}\cap \text{B }\\!\\!|\\!\\!\text{ + }\\!\\!|\\!\\!\text{ B}\cap \text{C }\\!\\!|\\!\\!\text{ + }\\!\\!|\\!\\!\text{ A}\cap \text{C }\\!\\!|\\!\\!\text{ = 14 +15 + 12 = 41}$.

So, now, the total number of members will be $76-41=35$. But while subtracting the number of members who play two games, we have also removed the number of members who play all three games. Therefore, we will add $|\text{A}\cap \text{B}\cap \text{C}|\text{ = 8}$ to the number of members we counted earlier. So, we will get $35+8=43$.

Hence, there are 43 members in total.

Note: We can solve this type of question using another method, which is Venn diagrams. The vein diagram for this question will look like the following,

In this diagram, the circles represent the three sets. It is easier to see the intersections of the sets in Venn diagrams. The numbers inside the regions in the diagram are the size of the set represented by that region.