In a class of 30 students 12 take needle work, 16 take physics and 18 take history. If all the 30 students take at least one subject and no one takes all three, then the number of students taking 2 subjects is

Options

A. 16
B. 6
C. 8
D. 20

Explanation

To solve the problem, we will use the principle of inclusion-exclusion to find the number of students taking exactly two subjects.

### Step-by-Step Solution:

1. **Identify the Variables:**
- Let \( N \) be the number of students taking Needlework = 12
- Let \( P \) be the number of students taking Physics = 16
- Let \( H \) be the number of students taking History = 18
- Let \( x \) be the number of students taking exactly two subjects.
- Total number of students = 30

2. **Set Up the Inclusion-Exclusion Principle:**
According to the principle of inclusion-exclusion for three sets, we have:
\[ |N \cup P \cup H| = |N| + |P| + |H| - |N \cap P| - |P \cap H| - |H \cap N| + |N \cap P \cap H| \]
Since no student is taking all three subjects, \( |N \cap P \cap H| = 0 \).

3. **Substituting the Values:**
We know that:
\[ |N \cup P \cup H| = 30 \]
Substituting the known values:
\[ 30 = 12 + 16 + 18 - (|N \cap P| + |P \cap H| + |H \cap N|) \]
Let \( a = |N \cap P| \), \( b = |P \cap H| \), \( c = |H \cap N| \). Then we can rewrite the equation as:
\[ 30 = 46 - (a + b + c) \]

4. **Rearranging the Equation:**
Rearranging gives:
\[ a + b + c = 46 - 30 = 16 \]

5. **Finding the Number of Students Taking Exactly Two Subjects:**
The number of students taking exactly two subjects can be calculated as:
\[ x = a + b + c \]
Since we found \( a + b + c = 16 \), we conclude that:
\[ x = 16 \]

### Final Answer:
The number of students taking exactly two subjects is **16**.