Question

In a group of students, 10 students like Mathematics, 12 students like English, 4 students like both Mathematics and English, and 6 students like neither Mathematics nor English. The number of students in the group is

• A. 18
• B. 20
• C. 24
• D. 32

Hint:

Use $|M \cup E| = |M|+|E|-|M \cap E|$ to find how many like at least one subject, then add the "neither" count for the total group size.

Answer 1

The Correct Option is C

Step 1: Set up the sets.
Let M be the set of students who like Mathematics and E be the set of students who like English. We are given \(|M|=10\), \(|E|=12\), \(|M \cap E| = 4\), and the number of students who like neither is 6.

Step 2: Apply the inclusion-exclusion principle.
The number of students who like Mathematics or English (or both) is
\[ |M \cup E| = |M| + |E| - |M \cap E| \]
This formula avoids double counting the students who like both subjects.

Step 3: Substitute the numbers.
\[ |M \cup E| = 10 + 12 - 4 = 18 \]
So 18 students like at least one of the two subjects.

Step 4: Add the students who like neither.
The total group size equals the students who like at least one subject plus the students who like neither.
\[ \text{Total} = |M \cup E| + \text{neither} = 18 + 6 = 24 \]

Step 5: Rule out the other options.
18 is only the count of students liking at least one subject, before adding the "neither" group, so it is incomplete. 20 and 32 do not come from a correct application of inclusion-exclusion on the given numbers.

Final Answer:
The group has 24 students. \[ \boxed{24} \]