Question

The number of ways in which a committee of 7 members can be formed from 6 teachers, 5 fathers and 4 students in such a way that at least one from each group is included and teachers form the majority among them, is:

• A. \(1865 \)
• B. \(2370 \)
• C. \(3050 \)
• D. \(4380 \)

Hint:

When dealing with "at least one" conditions and "majority" rules in committee formation, break down the problem into cases based on the number of members from the group forming the majority. Carefully list all combinations that satisfy the minimum requirements for each group and the specific majority definition (e.g., more than half of the total committee members, or the largest single group as interpreted here).

Answer 1

The Correct Option is B

Step 1: Understand the problem statement.
We need to form a committee of 7 members by selecting from three groups: Teachers (T), Fathers (F), and Students (S). The available members are:
- Teachers (T): 6
- Fathers (F): 5
- Students (S): 4
Step 2: Understand the conditions for committee formation.
Let \(n_T\), \(n_F\), and \(n_S\) be the number of teachers, fathers, and students selected, respectively.
The conditions are:
1. Total members: \(n_T + n_F + n_S = 7\).
2. At least one from each group: \(n_T \ge 1\), \(n_F \ge 1\), \(n_S \ge 1\).
3. Teachers form the majority among them: This means the number of teachers selected must be strictly greater than the number of fathers selected AND strictly greater than the number of students selected. So, \(n_T > n_F\) and \(n_T > n_S\).
Step 3: List all possible combinations of (\(n_T, n_F, n_S\)) that satisfy all conditions.
We need to find integer solutions for \(n_T + n_F + n_S = 7\) with \(1 \le n_T \le 6\), \(1 \le n_F \le 5\), \(1 \le n_S \le 4\), and \(n_T > n_F\), \(n_T > n_S\).
Case 1: \(n_T = 3\)
We need \(n_F + n_S = 7 - 3 = 4\).
Also \(n_F \ge 1\), \(n_S \ge 1\), and \(n_F < 3\), \(n_S < 3\).
Possible \((n_F, n_S)\) pairs satisfying \(n_F + n_S = 4\) and \(n_F < 3\), \(n_S < 3\):
- \((2, 2)\): \(n_T = 3\), \(n_F = 2\), \(n_S = 2\). Here \(3 > 2\) (satisfied).
Number of ways: \(C(6, 3) \times C(5, 2) \times C(4, 2) = 20 \times 10 \times 6 = 1200\).
Case 2: \(n_T = 4\)
We need \(n_F + n_S = 7 - 4 = 3\).
Also \(n_F \ge 1\), \(n_S \ge 1\), and \(n_F < 4\), \(n_S < 4\).
Possible \((n_F, n_S)\) pairs satisfying \(n_F + n_S = 3\) and \(n_F < 4\), \(n_S < 4\):
- \((1, 2)\): \(n_T = 4\), \(n_F = 1\), \(n_S = 2\). Here \(4 > 1\) and \(4 > 2\) (satisfied).
Number of ways: \(C(6, 4) \times C(5, 1) \times C(4, 2) = 15 \times 5 \times 6 = 450\).
- \((2, 1)\): \(n_T = 4\), \(n_F = 2\), \(n_S = 1\). Here \(4 > 2\) and \(4 > 1\) (satisfied).
Number of ways: \(C(6, 4) \times C(5, 2) \times C(4, 1) = 15 \times 10 \times 4 = 600\).
Total for \(n_T = 4\) cases: \(450 + 600 = 1050\).
Case 3: \(n_T = 5\)
We need \(n_F + n_S = 7 - 5 = 2\).
Also \(n_F \ge 1\), \(n_S \ge 1\), and \(n_F < 5\), \(n_S < 5\).
Possible \((n_F, n_S)\) pair satisfying \(n_F + n_S = 2\) and \(n_F < 5\), \(n_S < 5\):
\((1, 1)\): \(n_T = 5\), \(n_F = 1\), \(n_S = 1\). Here \(5 > 1\) (satisfied).
Number of ways: \(C(6, 5) \times C(5, 1) \times C(4, 1) = 6 \times 5 \times 4 = 120\).
\underline{Case 4: \(n_T = 6\)}
We need \(n_F + n_S = 7 - 6 = 1\).
Also \(n_F \ge 1\), \(n_S \ge 1\).
There are no pairs \((n_F, n_S)\) such that \(n_F \ge 1\) AND \(n_S \ge 1\) and \(n_F + n_S = 1\). (e.g., if \(n_F = 1\), then \(n_S = 0\), which violates \(n_S \ge 1\)). So, 0 ways for this case.
Step 4: Calculate the total number of ways.
Sum the ways from all valid cases:
Total ways = (Ways for \(n_T = 3\)) + (Ways for \(n_T = 4\)) + (Ways for \(n_T = 5\))
Total ways = \(1200 + 1050 + 120 = 2370\).
The final answer is \(\boxed{2370}\).