Question

If six persons are to be selected to form a committee from a group of seven women and four men so that at least three women are there on the committee, then the number of ways it can be done, is

• A. 641
• B. 545
• C. 591
• D. 441
• E. 139

Hint:

Logic Tip: A faster way using complements: Calculate total unrestricted combinations ($^{11}C_6 = 462$) and subtract the invalid cases. The only invalid cases are those with fewer than 3 women: 2W/4M ($^7C_2 \cdot ^4C_4 = 21 \cdot 1 = 21$). $462 - 21 = 441$.

Answer 1

The Correct Option is D

Concept:
When a combination has an "at least" condition, break it down into all possible valid sub-cases. The total number of ways is the sum of the combinations calculated for each distinct case. We need a committee of 6 people from 7 Women (W) and 4 Men (M), with $\ge 3$ W.
Step 1: Identify all valid sub-cases.
Case 1: 3 Women and 3 Men (Total = 6) Case 2: 4 Women and 2 Men (Total = 6) Case 3: 5 Women and 1 Man (Total = 6) Case 4: 6 Women and 0 Men (Total = 6)
Step 2: Calculate combinations for each case.
Case 1: $^7C_3 \cdot ^4C_3$ $$= \left(\frac{7 \cdot 6 \cdot 5}{3 \cdot 2 \cdot 1}\right) \cdot 4 = 35 \cdot 4 = 140$$ Case 2: $^7C_4 \cdot ^4C_2$ (Note that $^7C_4 = ^7C_3 = 35$) $$= 35 \cdot \left(\frac{4 \cdot 3}{2 \cdot 1}\right) = 35 \cdot 6 = 210$$ Case 3: $^7C_5 \cdot ^4C_1$ $$= \left(\frac{7 \cdot 6}{2 \cdot 1}\right) \cdot 4 = 21 \cdot 4 = 84$$ Case 4: $^7C_6 \cdot ^4C_0$ $$= 7 \cdot 1 = 7$$
Step 3: Sum the combinations of all valid cases.
$$\text{Total ways} = 140 + 210 + 84 + 7 = 441$$