Finding Students Who Like Only Computer

The problem asks us to determine the number of students who like only Computer given information about student preferences for Maths, English, and Computer.

Given Information Summary

We are provided with the following details about a class of 65 students:

• Total students: 65
• Students liking only Maths: 20
• Students liking only English: 25
• Students liking Maths and English: 15
• Total students liking Computer: 8
• Students liking all three subjects (Maths, English, Computer): 3
• Students liking Computers and English: 0
• Students liking Maths and Computers: 0

Analysis of Computer Subject Preferences

Let M represent Maths, E represent English, and C represent Computer.

We are given key constraints:

• The number of students liking Maths and Computers is zero: $ |M \cap C| = 0 $.
• The number of students liking Computers and English is zero: $ |C \cap E| = 0 $.

These constraints imply that there are no students who like Maths and Computer together (regardless of English preference), and no students who like English and Computer together (regardless of Maths preference).

Specifically, the number of students liking exactly two subjects involving Computer is zero:

• Students liking Maths and Computer only: $ |M \cap C \setminus E| = 0 $.
• Students liking English and Computer only: $ |E \cap C \setminus M| = 0 $.

Calculating Only Computer Students

The total number of students who like Computer ($ |C| $) is composed of four distinct groups:

1. Students liking only Computer ($ |C \setminus (M \cup E)| $).
2. Students liking Maths and Computer only ($ |M \cap C \setminus E| $).
3. Students liking English and Computer only ($ |E \cap C \setminus M| $).
4. Students liking all three subjects ($ |M \cap E \cap C| $).

We can write this as an equation:

Substituting the known values:

• $ |C| = 8 $
• $ |M \cap C \setminus E| = 0 $
• $ |E \cap C \setminus M| = 0 $
• $ |M \cap E \cap C| = 3 $

Plugging these into the equation:

Now, we solve for the number of students who like only Computer ($ |C \setminus (M \cup E)| $):

Therefore, 5 students like only Computer.

Verification (Optional)

We can verify this by calculating the total number of students accounted for:

• Only Maths: 20
• Only English: 25
• Only Computer: 5
• Maths and English only ($ |M \cap E| - |M \cap E \cap C| = 15 - 3 $): 12
• Maths and Computer only: 0
• English and Computer only: 0
• All three subjects: 3

Total accounted for = $ 20 + 25 + 5 + 12 + 0 + 0 + 3 = 65 $. This matches the total number of students in the class.