Students are in clubs as follows: Science -20, Drama - 30,and Band - 12. No student is in all three clubs, but 8 are in both Science and Drama, 6 are in both Science and Band, and 4 are in Drama and Band. How many different students are in at least on of the three clubs?

Answer

To solve the problem step by step, we can use the principle of inclusion-exclusion for three sets. Let's denote:

- \( N_S \): Number of students in the Science club = 20
- \( N_D \): Number of students in the Drama club = 30
- \( N_B \): Number of students in the Band club = 12
- \( N_{SD} \): Number of students in both Science and Drama = 8
- \( N_{SB} \): Number of students in both Science and Band = 6
- \( N_{DB} \): Number of students in both Drama and Band = 4
- \( N_{SDB} \): Number of students in all three clubs = 0 (given)

We need to find the total number of different students in at least one of the three clubs, which is represented as \( N_{S \cup D \cup B} \).

### Step 1: Apply the Inclusion-Exclusion Principle
The formula for the union of three sets is given by:

\[ N_{S \cup D \cup B} = N_S + N_D + N_B - N_{SD} - N_{SB} - N_{DB} + N_{SDB} \]

### Step 2: Substitute the Values
Now, we can substitute the known values into the formula:

\[ N_{S \cup D \cup B} = 20 + 30 + 12 - 8 - 6 - 4 + 0 \]

### Step 3: Calculate the Total
Now, let's perform the calculations step by step:

1. Calculate the sum of students in each club:
\[ 20 + 30 + 12 = 62 \]

2. Calculate the sum of students in the intersections:
\[ 8 + 6 + 4 = 18 \]

3. Now, substitute these sums back into the equation:
\[ N_{S \cup D \cup B} = 62 - 18 + 0 = 44 \]

### Conclusion
Thus, the total number of different students in at least one of the three clubs is:

\[ \boxed{44} \]