**Topic**

Class 10A has 27 students participating in at least one of two football and chess clubs, of which 19 students join the football club, 15 students join the chess club.

a) How many students are in the football club but not in the chess club?

b) How many students join both clubs?

c) Know that there are 8 students in the class who do not join any of the two clubs above. How many students are there in class 10A?

**Detailed explanation**

Let A be the set of students participating in the football club,

B is the set of students who join the chess club.

\( \Rightarrow A \cup B\) is the set of students who join at least one of the two football and chess clubs

\(A \cap B\) is the set of students who join both clubs

a) We have: \(n\left( A \right) = 19,{\rm{ }}n\left( B \right) = 15,n(A \cup B) = 27\)

The set of students who joined the football club but did not join the chess club is the set A\B (orange part) or the set \((A \cup B)\backslash B\)

The number of elements of the set \((A \cup B)\backslash B\) is the number of students in one of the two clubs minus the number of students in the chess club.

\(n((A \cup B)\backslash B) = n(A \cup B)–n\left( B \right) = 27–15 = 12.\)

So the number of students who join the football club but not the chess club is 12 students.

b) The set of students participating in both clubs is the set \(A \cap B\).

The number of elements of the set \(A \cap B\) is equal to the number of students in the football club minus the number of students who only join the football club but not the chess club.

\( \Rightarrow n(A \cap B) = n\left( A \right)–n(A{\rm{\backslash }}B) = 19–12 = 7.\)

So the number of students participating in both clubs is 7 students

c) Grade 10A consists of students who join at least one of the two clubs and students who do not.

=> number of students in class 10A is: 27 + 8 = 35 (students)

So the number of students in class 10A is 35 students.