Question

From a group of 5 men and 7 women, five persons are to be selected to form a committee so that at least 3 men are there in the committee. The number of ways this can be done is

• A. 181
• B. 274
• C. 246
• D. None of these

Hint:

In selection problems, break the problem down into cases based on conditions like "at least" and calculate the number of ways for each case.

Answer 1

The Correct Option is C

Step 1: Identify possible cases.
We need to select 5 persons with at least 3 men in the committee. We can have the following cases: - Case 1: 3 men and 2 women - Case 2: 4 men and 1 woman - Case 3: 5 men and 0 women
Step 2: Calculate number of ways for each case.
- Case 1: Number of ways to select 3 men from 5 and 2 women from 7: \[ \binom{5}{3} \times \binom{7}{2} = 10 \times 21 = 210 \] - Case 2: Number of ways to select 4 men from 5 and 1 woman from 7: \[ \binom{5}{4} \times \binom{7}{1} = 5 \times 7 = 35 \] - Case 3: Number of ways to select 5 men from 5: \[ \binom{5}{5} = 1 \]
Step 3: Total number of ways.
Total = 210 + 35 + 1 = 246.
Step 4: Conclusion.
Thus, the number of ways to form the committee is 246, which corresponds to option (C).

Answer 2

Instead of adding up the three qualifying cases directly, let's use complementary counting: find the total number of 5-person committees from all 12 people, then subtract the committees that fail to have at least 3 men (i.e. those with 0, 1, or 2 men).

Total committees of 5 from 12 people: \( \binom{12}{5} = 792 \). Committees with fewer than 3 men: \[ \binom{5}{0}\binom{7}{5} + \binom{5}{1}\binom{7}{4} + \binom{5}{2}\binom{7}{3} = 1\times21 + 5\times35 + 10\times35 = 21+175+350 = 546. \] Subtracting, the number of committees with at least 3 men is \( 792 - 546 = 246 \).

1. 181: This does not match either the direct case-sum or the complementary count above, so it is incorrect.
2. 274: This also does not agree with \( 792-546 \), so it is incorrect.
3. 246: This matches the complementary-counting result exactly.
4. None of these: Since \( 246 \) is obtainable and listed as option (C), this does not apply.

The complement of "fewer than 3 men" out of all possible committees reproduces the same count as adding the three qualifying cases directly.

Therefore, the correct answer is 246.