From a group of six men and four ladies (a total of ten people), a committee of three members is to be formed. Mrs X is not willing to join any committee in which Mr Y is a member, whereas Mr Y is willing to join a committee only if Mrs Z is also included. Under these conditions, how many different committees of three members are possible?

Introduction / Context:
This is a conditional combinations problem involving three specific people with constraints on which of them can serve together on a committee. It tests the ability to interpret verbal restrictions correctly and to translate them into mathematical conditions. Such questions are common in reasoning sections as they combine logical conditions with combinatorial counting.

Given Data / Assumptions:

• Total people: 6 men and 4 ladies, for a total of 10 individuals.
• Special persons: Mrs X, Mrs Z and Mr Y are among these ten.
• A committee of 3 members is to be formed.
• Mrs X will not join any committee that includes Mr Y.
• Mr Y will join a committee only if Mrs Z is also on that committee.
• We assume that Mrs X and Mrs Z are different ladies, and Mr Y is one of the men.

Concept / Approach:
We break the counting into cases based on whether Mr Y is in the committee or not. When Y is included, both conditions involving X and Z must be enforced. When Y is excluded, only the first condition becomes irrelevant and the committee selection is unrestricted. We then add the counts from the disjoint cases to get the total number of valid committees.

Step-by-Step Solution:
Step 1: Case 1: Committees that include Mr Y. Since Y is in the committee, Mrs Z must also be in the committee, and Mrs X cannot be included.Step 2: When Y is in, Z must be in, and X must be out. So two members are fixed: Y and Z. We need one more member from the remaining people (excluding X, Y and Z).Step 3: There are 10 total people minus these 3, leaving 7 people who can fill the third spot. So the number of committees in Case 1 is 7C1 = 7.Step 4: Case 2: Committees that do not include Mr Y. Then the conditions about X and Z involving Y do not apply, and we can choose any 3 people from the remaining 9 individuals (since Y is excluded).Step 5: Number of such committees is 9C3 = (9 × 8 × 7) / (3 × 2 × 1) = 84.Step 6: Total valid committees = committees with Y + committees without Y = 7 + 84 = 91.
Verification / Alternative check:
We can run a mental check: there are 10C3 = 120 possible committees without restrictions. We exclude those that violate the conditions. The derived count 91 means 29 committees are invalid. Quick reasoning shows that invalid committees are exactly those containing Y without Z or those containing both X and Y together. Counting these invalid combinations separately also leads to 29 and confirms the total of 91 valid committees.

Why Other Options Are Wrong:
Values like 98, 104, or 109 arise from miscounting one of the cases or forgetting the exclusion of X when Y is present. Only 91 is consistent with a careful case based analysis and with subtracting invalid committees from the unrestricted total of 120.

Common Pitfalls:
Students often misinterpret the statement about Y joining only if Z is included and may treat it as a biconditional (Z must also always be with Y) instead of a one-way condition. Another error is forgetting that X is prohibited whenever Y is present. Clearly separating the cases based on whether Y is in or out of the committee helps prevent such logical errors.

Final Answer:
The number of different committees that can be formed under the given conditions is 91.