There are 100 students in a class. In the examination, 50 of them failed in Mathematics, 45 failed in Physics, 40 failed in Biology and 32 failed in exactly two of the three subjects. Only one student passed in all the subjects. Then, the number of students failing in all the three subjects is

Options

A. 12
B. 4
C. 2
D. Cannot be determined

Explanation

To solve the problem, we will use the principle of inclusion-exclusion and the information provided about the students failing in different subjects.

### Step-by-Step Solution:

1. **Define the Sets**:
Let:
- \( M \) = Set of students failing in Mathematics
- \( P \) = Set of students failing in Physics
- \( B \) = Set of students failing in Biology

Given:
- \( |M| = 50 \)
- \( |P| = 45 \)
- \( |B| = 40 \)
- \( |M \cap P \cap B| = x \) (students failing in all three subjects)
- \( |M \cap P| + |P \cap B| + |B \cap M| - 3x = 32 \) (students failing in exactly two subjects)
- Only 1 student passed all subjects, so \( |M \cup P \cup B| = 100 - 1 = 99 \).

2. **Use the Inclusion-Exclusion Principle**:
According to the principle of inclusion-exclusion:
\[ |M \cup P \cup B| = |M| + |P| + |B| - |M \cap P| - |P \cap B| - |B \cap M| + |M \cap P \cap B| \]
Substituting the known values:
\[ 99 = 50 + 45 + 40 - (|M \cap P| + |P \cap B| + |B \cap M|) + x \]

3. **Simplify the Equation**:
\[ 99 = 135 - (|M \cap P| + |P \cap B| + |B \cap M|) + x \]
Rearranging gives:
\[ |M \cap P| + |P \cap B| + |B \cap M| = 135 - 99 + x = 36 + x \]

4. **Set Up the Equation for Exactly Two Subjects**:
From the previous step, we have:
\[ |M \cap P| + |P \cap B| + |B \cap M| - 3x = 32 \]
Substitute \( |M \cap P| + |P \cap B| + |B \cap M| \) from the earlier equation:
\[ (36 + x) - 3x = 32 \]
Simplifying this gives:
\[ 36 - 2x = 32 \]
\[ 2x = 4 \implies x = 2 \]

5. **Conclusion**:
The number of students failing in all three subjects is \( x = 2 \).

### Final Answer:
The number of students failing in all three subjects is **2**.