In a class, 18 students took Physics, 23 students took Chemistry and 24 students took Mathematics of these 13 took both Chemistry and Mathematics, 12 took both Physics and Chemistry and 11 took both Physics and Mathematics. If 6 students offered all the three subjects, find:
(i) The total number of students.

Answer

To solve the problem step by step, we will use the principle of inclusion-exclusion to find the total number of students in the class.

### Step 1: Define the Variables
Let:
- \( n(P) \) = Number of students who took Physics = 18
- \( n(C) \) = Number of students who took Chemistry = 23
- \( n(M) \) = Number of students who took Mathematics = 24
- \( n(C \cap M) \) = Number of students who took both Chemistry and Mathematics = 13
- \( n(P \cap C) \) = Number of students who took both Physics and Chemistry = 12
- \( n(P \cap M) \) = Number of students who took both Physics and Mathematics = 11
- \( n(P \cap C \cap M) \) = Number of students who took all three subjects = 6

### Step 2: Apply the Inclusion-Exclusion Principle
The total number of students who took at least one of the subjects can be calculated using the formula:

\[ n(P \cup C \cup M) = n(P) + n(C) + n(M) - n(P \cap C) - n(C \cap M) - n(P \cap M) + n(P \cap C \cap M) \]

### Step 3: Substitute the Values
Now, substituting the values we defined in Step 1 into the formula:

\[ n(P \cup C \cup M) = 18 + 23 + 24 - 12 - 13 - 11 + 6 \]

### Step 4: Perform the Calculations
Now, let's do the arithmetic step by step:

1. Calculate the sum of students who took each subject:
\[ 18 + 23 + 24 = 65 \]

2. Calculate the sum of students who took two subjects:
\[ 12 + 13 + 11 = 36 \]

3. Now, substitute these sums back into the equation:
\[ n(P \cup C \cup M) = 65 - 36 + 6 \]

4. Calculate:
\[ 65 - 36 = 29 \]
\[ 29 + 6 = 35 \]

### Final Answer
Thus, the total number of students in the class is **35**.

---