Question

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 \(100k\), then k is equal to __________.

Hint:

Break the selection into mutually exclusive cases based on the sum constraint. This reduces the problem to smaller, manageable counting tasks.

Answer 1

Correct Answer: 238

Step 1: Understanding the Concept:
This is a problem of selection (combinations) with constraints.
We must identify all valid distributions of students among the three classes that satisfy both the minimum per-class requirement and the group sum limit.
Step 3: Detailed Explanation:
Let the number of students selected from classes 10, 11, and 12 be \(x_{10}, x_{11},\) and \(x_{12}\).
Given:
1. \(x_{10} + x_{11} + x_{12} = 10\).
2. \(x_{10} \ge 2, x_{11} \ge 2, x_{12} \ge 2\).
3. \(x_{10} + x_{11} \le 5\).
From (2), \(x_{10} + x_{11} \ge 2 + 2 = 4\).
Combined with (3), \(x_{10} + x_{11}\) can only be 4 or 5.

Case 1: \(x_{10} + x_{11} = 4\)
This implies \(x_{12} = 10 - 4 = 6\).
The only solution for \(x_{10}, x_{11} \ge 2\) is \((2, 2)\).
Ways \(= \binom{5}{2} \times \binom{6}{2} \times \binom{8}{6} = 10 \times 15 \times 28 = 4200\).

Case 2: \(x_{10} + x_{11} = 5\)
This implies \(x_{12} = 10 - 5 = 5\).
Solutions for \((x_{10}, x_{11})\): \((2, 3)\) and \((3, 2)\).
Ways for (2, 3, 5) \(= \binom{5}{2} \times \binom{6}{3} \times \binom{8}{5} = 10 \times 20 \times 56 = 11200\).
Ways for (3, 2, 5) \(= \binom{5}{3} \times \binom{6}{2} \times \binom{8}{5} = 10 \times 15 \times 56 = 8400\).

Total ways \(= 4200 + 11200 + 8400 = 23800\).
Given Total \(= 100k \implies k = 238\).
Step 4: Final Answer:
The value of \(k\) is 238.