Question

In a class of 28 pupils, 13 have pencils, 9 have erasers and 9 have neither pencils not erasers. How many pupils have both pencils and erasers ?

Answer 1

To find out how many pupils have both pencils and erasers, we can use the inclusion-exclusion principle.

Let:
- \( P \) be the set of pupils who have pencils.
- \( E \) be the set of pupils who have erasers.

From the information given:
- The total number of pupils \( |U| = 28 \)
- The number of pupils with pencils \( |P| = 13 \)
- The number of pupils with erasers \( |E| = 9 \)
- The number of pupils with neither pencils nor erasers = 9

First, we can find the number of pupils who have either pencils, erasers, or both. This is given by:

\[
|U| - \text{(number of pupils with neither)} = 28 - 9 = 19
\]

Thus, \( |P \cup E| = 19 \) (the number of pupils who have either pencils or erasers or both).

According to the principle of inclusion-exclusion, we have:

\[
|P \cup E| = |P| + |E| - |P \cap E|
\]

Where \( |P \cap E| \) is the number of pupils who have both pencils and erasers.

Plugging in the values we have:

\[
19 = 13 + 9 - |P \cap E|
\]

Simplifying the right-hand side:

\[
19 = 22 - |P \cap E|
\]

This gives us:

\[
|P \cap E| = 22 - 19
\]
\[
|P \cap E| = 3
\]

Therefore, the number of pupils who have both pencils and erasers is \( \boxed{3} \).