There are 5 students in class 10, 6 students in class 11 and 8 students in class 12. If the number of ways, in which 10 students can be selected from them so as to include at least 2 students from each class and at most 5 students from the total 11 students of class 10 and 11 is 100 k, then k is equal to _______.

Options

A. 250
B. 238
C. 150
D. 338

Answer

To solve the problem, we need to select 10 students from three classes (10, 11, and 12) with specific constraints. Let's break down the solution step by step.

### Step 1: Understand the constraints
- We have 5 students in class 10, 6 students in class 11, and 8 students in class 12.
- We need to select 10 students in total.
- We must include at least 2 students from each class.
- We can select at most 5 students from the combined total of classes 10 and 11.

### Step 2: Determine the possible distributions
Given that we need at least 2 students from each class, we can denote:
- \( x_{10} \): Number of students selected from class 10
- \( x_{11} \): Number of students selected from class 11
- \( x_{12} \): Number of students selected from class 12

From the problem, we have:
1. \( x_{10} + x_{11} + x_{12} = 10 \)
2. \( x_{10} \geq 2 \)
3. \( x_{11} \geq 2 \)
4. \( x_{12} \geq 2 \)
5. \( x_{10} + x_{11} \leq 5 \)

### Step 3: Set up cases based on the distribution
Since we need at least 2 from each class, we can start by allocating 2 students to each class:
- This gives us \( x_{10} = 2 + a \), \( x_{11} = 2 + b \), \( x_{12} = 2 + c \) where \( a, b, c \) are non-negative integers.
- The equation becomes: \( (2 + a) + (2 + b) + (2 + c) = 10 \) or \( a + b + c = 4 \).

Now, we also have the constraint \( (2 + a) + (2 + b) \leq 5 \), which simplifies to \( a + b \leq 1 \).

### Step 4: Analyze the cases
#### Case 1: \( a + b = 0 \)
- This means \( a = 0 \) and \( b = 0 \).
- Thus, \( x_{10} = 2 \), \( x_{11} = 2 \), and \( x_{12} = 6 \).
- The number of ways to choose:
\[ \binom{5}{2} \cdot \binom{6}{2} \cdot \binom{8}{6} \]

#### Case 2: \( a + b = 1 \)
- Possible combinations are \( (a, b) = (1, 0) \) or \( (0, 1) \).
- For \( (1, 0) \): \( x_{10} = 3 \), \( x_{11} = 2 \), \( x_{12} = 5 \).
\[ \binom{5}{3} \cdot \binom{6}{2} \cdot \binom{8}{5} \]
- For \( (0, 1) \): \( x_{10} = 2 \), \( x_{11} = 3 \), \( x_{12} = 5 \).
\[ \binom{5}{2} \cdot \binom{6}{3} \cdot \binom{8}{5} \]

### Step 5: Calculate the total number of ways
Now, we can calculate the total number of ways for each case:
1. **Case 1**:
\[ \binom{5}{2} = 10, \quad \binom{6}{2} = 15, \quad \binom{8}{6} = 28 \]
Total for Case 1:
\[ 10 \cdot 15 \cdot 28 = 4200 \]

2. **Case 2**:
- For \( (1, 0) \):
\[ \binom{5}{3} = 10, \quad \binom{6}{2} = 15, \quad \binom{8}{5} = 56 \]
Total for \( (1, 0) \):
\[ 10 \cdot 15 \cdot 56 = 8400 \]
- For \( (0, 1) \):
\[ \binom{5}{2} = 10, \quad \binom{6}{3} = 20, \quad \binom{8}{5} = 56 \]
Total for \( (0, 1) \):
\[ 10 \cdot 20 \cdot 56 = 11200 \]

### Step 6: Sum all cases
Total ways:
\[ 4200 + 8400 + 11200 = 28800 \]

### Step 7: Relate to the given equation
We know that the total number of ways is equal to \( 100k \):
\[ 100k = 28800 \implies k = \frac{28800}{100} = 288 \]

### Final Answer
Thus, the value of \( k \) is \( \boxed{288} \).