Q.In a class of 60 students, 25 students play cricket and 20 students play tennis, and 10 students play both the games. Find the number of students who play neither?

The key idea here is Set Operations, specifically using the Principle of Inclusion-Exclusion to find the union of two sets and then its complement.

1. Let $C$ be the set of students who play cricket and $T$ be the set of students who play tennis. The total number of students is $|U| = 60$.
2. We are given $|C| = 25$, $|T| = 20$, and $|C \cap T| = 10$.
3. The number of students who play at least one game is given by the Principle of Inclusion-Exclusion: $|C \cup T| = |C| + |T| - |C \cap T| = 25 + 20 - 10 = 35$. …

When dealing with groups of people or objects that have certain characteristics, and some individuals share multiple characteristics, set theory provides a clear framework. Here, we have a group of students, and some play cricket, some play tennis, and some play both. Our goal is to find those who play neither.

The core idea is to first figure out how many students play at least one of the games. Once we have this number, we can subtract it from the total number of students in the class to find those who play none of the games.

Let's define our sets:

• Let $U$ be the set of all students in the class. So, $n(U) = 60$.
• Let $C$ be the set of students who play cricket. So, $n(C) = 25$.
• Let $T$ be the set of students who play tennis. So, $n(T) = 20$.
• The set of students who play both games is the intersection of $C$ and $T$, denoted $C \cap T$. So, $n(C \cap T) = 10$.

We are looking for the number of students who play neither game. This corresponds to the complement of the union of $C$ and $T$ with respect to $U$, i.e., $n((C \cup T)')$.

1. Understand the Principle of Inclusion-Exclusion:

   When we add the number of students who play cricket ($n(C)$) and the number of students who play tennis ($n(T)$), we are counting the students who play both games twice. Once as part of the cricket players and once as part of the tennis players. To get the total number of students who play at least one game (the union $C \cup T$), we must subtract the number of students who play both games ($n(C \cap T)$) once. This ensures that those who play both are counted exactly once.

   For any two finite sets $A$ and $B$, the number of elements in their union is given by:

   $n(A \cup B) = n(A) + n(B) - n(A \cap B)$

2. Calculate the number of students who play at least one game:

   Using the Principle of Inclusion-Exclusion for our sets $C$ and $T$:

   $n(C \cup T) = n(C) + n(T) - n(C \cap T)$

   Substitute the given values:

   $n(C \cup T) = 25 + 20 - 10$

   $n(C \cup T) = 45 - 10$

   $n(C \cup T) = 35$

   This means 35 students play either cricket, or tennis, or both.

   Tip

   You can also visualize this with a Venn diagram.

   • Students playing only cricket: $n(C) - n(C \cap T) = 25 - 10 = 15$ …