To solve the problem of how many ways a student can solve exactly 5 questions from three sections (A, B, and C), with the condition that at least one question must be taken from each section, we can break it down into steps.
### Step-by-Step Solution:
1. **Understanding the Problem**:
We have three sections (A, B, and C), each with 5 questions. We need to select exactly 5 questions such that at least one question is chosen from each section.
2. **Distributing the Questions**:
Since we need to select exactly 5 questions and at least one from each section, we can represent the distribution of questions from each section as follows:
- Let \( x_A \) be the number of questions chosen from section A,
- Let \( x_B \) be the number of questions chosen from section B,
- Let \( x_C \) be the number of questions chosen from section C.
The equation we need to satisfy is:
\[
x_A + x_B + x_C = 5
\]
with the constraints \( x_A \geq 1 \), \( x_B \geq 1 \), and \( x_C \geq 1 \).
3. **Transforming the Variables**:
To simplify the equation, we can set:
\[
y_A = x_A - 1, \quad y_B = x_B - 1, \quad y_C = x_C - 1
\]
This transforms our equation into:
\[
(y_A + 1) + (y_B + 1) + (y_C + 1) = 5
\]
which simplifies to:
\[
y_A + y_B + y_C = 2
\]
where \( y_A, y_B, y_C \geq 0 \).
4. **Finding Non-Negative Solutions**:
The number of non-negative integer solutions to the equation \( y_A + y_B + y_C = 2 \) can be found using the "stars and bars" theorem. The formula for the number of solutions is given by:
\[
\binom{n+k-1}{k-1}
\]
where \( n \) is the total number of items to distribute (2 in this case) and \( k \) is the number of sections (3 here). Thus, we have:
\[
\binom{2+3-1}{3-1} = \binom{4}{2} = 6
\]
5. **Calculating the Ways to Choose Questions**:
Now we need to consider the different distributions of questions:
- **Case 1**: (1, 1, 3) - One question from two sections and three from one section.
- **Case 2**: (1, 2, 2) - One question from one section and two from the other two sections.
For each case, we calculate the number of ways to choose questions from each section.
- **Case 1**:
- (1, 1, 3): Choose 1 from A, 1 from B, and 3 from C:
\[
\text{Ways} = \binom{5}{1} \cdot \binom{5}{1} \cdot \binom{5}{3} = 5 \cdot 5 \cdot 10 = 250
\]
- Similarly, for (1, 3, 1) and (3, 1, 1), we have:
\[
250 \text{ (for each arrangement)} \Rightarrow 3 \cdot 250 = 750
\]
- **Case 2**:
- (1, 2, 2): Choose 1 from A, 2 from B, and 2 from C:
\[
\text{Ways} = \binom{5}{1} \cdot \binom{5}{2} \cdot \binom{5}{2} = 5 \cdot 10 \cdot 10 = 500
\]
- Similarly, for (2, 1, 2) and (2, 2, 1), we have:
\[
500 \text{ (for each arrangement)} \Rightarrow 3 \cdot 500 = 1500
\]
6. **Total Ways**:
Finally, we add the results from both cases:
\[
\text{Total Ways} = 750 + 1500 = 2250
\]
### Final Answer:
The total number of ways a student can solve exactly 5 questions, taking at least one question from each section, is **2250**.
