How many students play basketball and soccer?

Question

A small school with 14 students has a basketball team with 5 players, a chess team with 4 players, a netball team with 7 players, and a soccer team with 11 players.

• Each student is on at least one of these teams and no student is on all four teams.

• Only one student, Hawa, is on three of the teams: she plays chess, netball and soccer.

• There is exactly 1 student who plays basketball and chess.

• There is exactly 1 student who plays basketball and netball.

• Hawa is the only student who plays chess and netball.

• There are exactly 3 students (including Hawa) who play chess and soccer.

• There are exactly 5 students (including Hawa) who play netball and soccer

How many students play basketball and soccer?

I did this manually and drew out the slots for each member in each team and I got 3 students.But I'm pretty sure there's a mathematical way of doing this using combinations perhaps.Can someone post the working?

Answer 1

You are correct.

We can verify your answer by using the Inclusion-Exclusion Principle.

Let $B$, $C$, $N$, and $S$ denote, respectively, the set of students who play basketball, chess, netball, and soccer.

If we apply the Inclusion-Exclusion Principle to these four sets, we obtain \begin{align*} |B \cup C \cup N \cup S| & = |B| + |C| + |N| + |S|\\ & \quad - |B \cap C| - |B \cap N| - |B \cap S| - |C \cap N| - |C \cap S| - |N \cap S|\\ & \quad + |B \cap C \cap N| + |B \cap C \cap S| + |B \cap N \cap S| - |C \cap N \cap S|\\ & \quad - |B \cap C \cap N \cap S|\\ \end{align*} We are told the following information:

• A small school with $14$ students has a basketball team with $5$ players $\implies |B| = 5$, a chess team with $4$ players $\implies |C| = 4$, a netball team with $7$ players $\implies |N| = 7$, and a soccer team with $11$ players $\implies |S| = 11$.
• Each student is on at least one of these teams $\implies |B \cup C \cup N \cup S| = 14$ and no student is on all four teams $\implies |B \cap C \cap N \cap S| = 0$.
• Only one student, Hawa, is on three of the teams: she plays chess, netball and soccer $\implies |C \cap N \cap S| = 1$ and $|B \cap C \cap N| = |B \cap C \cap S| = |B \cap N \cap S| = 0$.
• There is exactly 1 student who plays basketball and chess $\implies |B \cap C| = 1$.
• There is exactly 1 student who plays basketball and netball $\implies |B \cap N| = 1$.
• Hawa is the only student who plays chess and netball $\implies |C \cap N| = 1$.
• There are exactly 3 students (including Hawa) who play chess and soccer $\implies |C \cap S| = 3$.
• There are exactly 5 students (including Hawa) who play netball and soccer $\implies |N \cap S| = 5$.

Substituting this information into the above formula gives \begin{align*} 14 & = 5 + 4 + 7 + 11 - 1 - 1 - |B \cap S| - 1 - 3 - 5 + 0 + 0 + 0 + 1 - 0\\ 14 & = 17 - |B \cap S|\\ |B \cap S| & = 3 \end{align*} as you found.