A committee of 5 persons is to be formed from a group of 6 gentlemen and 4 ladies. In how many ways can this be done if the committee must include at least one lady?

Introduction / Context:
This question deals with forming a committee from a mixed group of gentlemen and ladies with a minimum representation constraint. We have to form a committee of 5 people from a group of 10 (6 gentlemen and 4 ladies), ensuring that at least one lady is included. This is a good example of applying combinations and using the complement method to handle an "at least one" condition efficiently.

Given Data / Assumptions:
- Gentlemen available: 6.

- Ladies available: 4.

- Total people: 6 + 4 = 10.

- Committee size: 5 members.

- Condition: the committee must contain at least one lady (1 or more).

- Order of members in the committee does not matter; we only care about the chosen set.

Concept / Approach:
We can count the number of valid committees by subtracting from the total number of possible committees (with no restrictions) the number of committees that violate the condition (that is, committees with zero ladies). The complement method is usually simpler than adding up cases for exactly 1 lady, exactly 2 ladies, and so on, which would require multiple combination terms.

Step-by-Step Solution:
Step 1: Compute the total number of ways to choose any 5 persons from the 10 available people, without any condition.Step 2: Total committees without restriction = 10C5.Step 3: Compute 10C5 = 252 (a standard value or via 10! / (5! * 5!) ).Step 4: Now count the number of committees with no ladies at all (only gentlemen).Step 5: Such committees must be formed entirely from the 6 gentlemen.Step 6: Number of all-gentlemen committees = 6C5 = 6.Step 7: Committees that contain at least one lady = total committees - all-gentlemen committees.Step 8: Required count = 10C5 - 6C5 = 252 - 6 = 246.

Verification / Alternative check:
As an alternative, we can sum over cases where the committee has exactly k ladies for k = 1, 2, 3, 4. For example, for exactly 1 lady, we choose 1 from 4 and 4 from 6 gentlemen; for exactly 2 ladies, choose 2 from 4 and 3 from 6, and so on. Adding all these cases yields the same result of 246, but it requires more arithmetic. The complement method is cleaner and less error-prone, and the equality of the two results verifies correctness.

Why Other Options Are Wrong:
- 123 and 113 are far too small and come from miscomputing either 10C5 or 6C5, or from trying to sum only some of the lady-count cases.

- 945 is much larger than 252 and cannot be correct for choosing 5 people out of 10; it reflects a serious misunderstanding of the combination formula.

Common Pitfalls:
A frequent error is to forget that committees with no ladies must be excluded and to simply compute 10C5. Another pitfall is incorrectly summing several overlapping cases for different numbers of ladies, leading to double-counting or missing some configurations. Carefully applying the complement method—total minus invalid—is the simplest strategy in most “at least one” problems.

Final Answer:
The committee can be formed in 246 different ways if it must include at least one lady.