Question

From a group of \(7\) men and \(6\) women, five persons are to be selected to form a committee so that at least \(3\) men are there on the committee. In how many ways can it be done?

• A. 564
• B. 645
• C. 735
• D. 756

Hint:

For “at least/at most” selection constraints, break into disjoint cases that exhaust all admissible splits (e.g., men–women counts), compute each with combinations, and sum.

Answer 1

The Correct Option is D

Step 1 (Interpret “at least 3 men”).
“At least \(3\) men” in a \(5\)-member committee means the admissible man–woman splits are: \((3\text{ men},2\text{ women})\), \((4,(a)\), and \((5,0)\).
Step 2 (Count each case using combinations).
Case A — \(3\) men and \(2\) women: ways \(= \binom{7}{3}\binom{6}{2}\).
Case B — \(4\) men and \(1\) woman: ways \(= \binom{7}{4}\binom{6}{1}\).
Case C — \(5\) men and \(0\) women: ways \(= \binom{7}{5}\binom{6}{0}\).
Step 3 (Compute each term).
\(\binom{7}{3}=35,\ \binom{6}{2}=15 35\times 15=525\).
\(\binom{7}{4}=35,\ \binom{6}{1}=6 35\times 6=210\).
\(\binom{7}{5}=21,\ \binom{6}{0}=1 21\times 1=21\).
Step 4 (Add the cases).
Total ways \(= 525 + 210 + 21 = 756\).
Step 5 (Conclusion).
Hence, the committee can be formed in \(756\) ways Option 4.
\[ \boxed{756\ \text{ways (Option (d)}} \]