In class 12 there are 8 students. In class 11 there are 6 students .In class 10 there are 5 students . Total ways of selecting 10 students , such that there are at least 2 students from each class and at most 5 students from 11 students of Class 10 and 11 combined is 100k. Then find value of k

Answer

To solve the problem, we need to find the total number of ways to select 10 students from three classes, ensuring that we meet the specified conditions. Let's break down the solution step by step.

### Step 1: Understand the Problem
We have:
- Class 12: 8 students
- Class 11: 6 students
- Class 10: 5 students

We need to select 10 students such that:
- At least 2 students from each class.
- At most 5 students from the combined total of Class 10 and Class 11.

### Step 2: Define the Cases
Since we need to select at least 2 students from each class, we can define two main cases based on how many students we select from Classes 10 and 11 combined:

1. **Case 1**: Select 4 students from Classes 10 and 11 combined (which means 6 from Class 12).
2. **Case 2**: Select 5 students from Classes 10 and 11 combined (which means 5 from Class 12).

### Step 3: Calculate for Case 1
In this case, we can select:
- 2 students from Class 10 and 2 students from Class 11.
- Therefore, we will select 6 students from Class 12.

The number of ways to do this is given by:
\[ \text{Ways} = \binom{8}{6} \times \binom{6}{2} \times \binom{5}{2} \]

### Step 4: Calculate for Case 2
In this case, we can select:
- 3 students from Class 10 and 2 students from Class 11.
- Therefore, we will select 5 students from Class 12.

The number of ways to do this is given by:
\[ \text{Ways} = \binom{8}{5} \times \binom{6}{2} \times \binom{5}{3} \]

We can also select:
- 2 students from Class 10 and 3 students from Class 11.
- Therefore, we will select 5 students from Class 12.

The number of ways to do this is given by:
\[ \text{Ways} = \binom{8}{5} \times \binom{6}{3} \times \binom{5}{2} \]

### Step 5: Combine the Cases
Now, we need to sum the total number of ways from both cases:
\[ \text{Total Ways} = \left( \binom{8}{6} \times \binom{6}{2} \times \binom{5}{2} \right) + \left( \binom{8}{5} \times \binom{6}{2} \times \binom{5}{3} \right) + \left( \binom{8}{5} \times \binom{6}{3} \times \binom{5}{2} \right) \]

### Step 6: Calculate Each Binomial Coefficient
Now we calculate each binomial coefficient:
- \(\binom{8}{6} = 28\)
- \(\binom{6}{2} = 15\)
- \(\binom{5}{2} = 10\)
- \(\binom{8}{5} = 56\)
- \(\binom{6}{3} = 20\)
- \(\binom{5}{3} = 10\)

### Step 7: Substitute and Calculate
Substituting the values we calculated:
\[ \text{Total Ways} = (28 \times 15 \times 10) + (56 \times 15 \times 10) + (56 \times 20 \times 10) \]

Calculating each term:
1. \(28 \times 15 \times 10 = 4200\)
2. \(56 \times 15 \times 10 = 8400\)
3. \(56 \times 20 \times 10 = 11200\)

Now summing these values:
\[ \text{Total Ways} = 4200 + 8400 + 11200 = 27800 \]

### Step 8: Relate to the Given Equation
According to the problem, the total number of ways is given as \(100k\):
\[ 100k = 27800 \]
Thus, solving for \(k\):
\[ k = \frac{27800}{100} = 278 \]

### Final Answer
The value of \(k\) is \(278\).