The Overlapping Committees

We need to check each pair's committee co-membership. Let's list committees per employee:

• Alice: X, Z
• Bob: X, Y
• Charlie: X, Y, W
• Dana: X, Z
• Evan: Y, W
• Fiona: Y
• George: Z
• Hannah: Z
• Ian: W
• Julia: W

Now check pairs that appear together in multiple committees:

• Alice & Dana: both in X and Z → share 2 committees.
• Bob & Charlie: both in X and Y → share 2 committees.
• Charlie & Evan: both in Y and W → share 2 committees.
• Others: Alice & Bob only in X; Alice & Charlie only in X; etc. So the pairs sharing more than one committee are: (Alice, Dana), (Bob, Charlie), (Charlie, Evan).

We need to choose 3 employees such that:

1. No pair shares more than one committee (so we cannot include any of the pairs: (Alice, Dana), (Bob, Charlie), (Charlie, Evan)).
2. At least one from X (Alice, Bob, Charlie, Dana) and at least one from Y (Bob, Charlie, Evan, Fiona).

We can enumerate all 3-person subsets from the 12 employees, but that's 220. Better to use inclusion-exclusion or systematic listing. Let's list all 3-person sets that meet the committee representation condition first, then exclude those with forbidden pairs.

Alternatively, we can consider the set of all employees, but note that some employees are only in one committee (Fiona, George, Hannah, Ian, Julia). They can be included as long as they don't create forbidden pairs (they won't, since they share at most one committee with anyone).

We need at least one from X and one from Y. Let's denote the sets: X = {A,B,C,D}, Y = {B,C,E,F}. The union X∪Y = {A,B,C,D,E,F}. Employees outside X∪Y are G,H,I,J (only in Z or W). They can be part of the task force but don't help satisfy the X and Y requirements.

We need to choose 3 from all 12, with at least one from X and one from Y.

Total 3-person sets with at least one from X and one from Y = total 3-person sets - sets with no X - sets with no Y + sets with no X and no Y.

Total 3-person sets from 12: C(12,3)=220. No X: choose from 8 employees (Y plus others not in X). But careful: Y has 4, others not in X are {E,F,G,H,I,J}? Actually X={A,B,C,D}. So not X = {E,F,G,H,I,J}? Wait, list all employees: A,B,C,D,E,F,G,H,I,J? That's 10? Let's recount: Alice, Bob, Charlie, Dana, Evan, Fiona, George, Hannah, Ian, Julia = 10 employees. I said 12 earlier, but the setup says 12 employees? The setup says "12 employees" but lists only 10 distinct initials. Let's check: Alice, Bob, Charlie, Dana, Evan, Fiona, George, Hannah, Ian, Julia. That's 10. Possibly there are two more not mentioned? The problem says 12 employees, but only 10 appear in committees. Maybe the other two are not on any committee? That would be odd. To be consistent, I'll assume the 12 employees are those listed plus two others not on any committee? But then they would be irrelevant. Better to treat the 10 as the employees. The problem might have a typo. I'll proceed with 10 employees. So total 3-person sets: C(10,3)=120.

No X: employees not in X: Y={B,C,E,F} plus others {G,H,I,J} = {B,C,E,F,G,H,I,J} that's 8. C(8,3)=56. No Y: employees not in Y: X={A,B,C,D} plus {G,H,I,J} = {A,B,C,D,G,H,I,J} that's 8. C(8,3)=56. No X and no Y: employees not in X and not in Y: {G,H,I,J} only? Also check: not in X: {B,C,E,F,G,H,I,J}; not in Y: {A,B,C,D,G,H,I,J}. Intersection: {G,H,I,J} (since B,C are in both X and Y, so they are in X and Y, so they are excluded from both no X and no Y? Actually "no X" means not in X, so B,C are in X, so they are not in no X. So intersection of not X and not Y is employees not in X and not in Y: that is {G,H,I,J} only. So C(4,3)=4. Thus sets with at least one X and one Y = 120 - 56 - 56 + 4 = 12.

Now from these 12, we must exclude those containing a forbidden pair: (A,D), (B,C), (C,E).

Let's list the 12 sets. We can list all 3-person sets from the 10 that have at least one from X and one from Y. Since X∪Y = {A,B,C,D,E,F} and others {G,H,I,J}. The condition requires at least one from X and one from Y. The 12 sets can be found by considering cases:

• Case 1: All 3 from X∪Y. Then we need at least one from X and one from Y. The number of such sets: total from X∪Y (6 employees) C(6,3)=20, minus sets with all from X (C(4,3)=4) minus sets with all from Y (C(4,3)=4) plus sets with all from X and all from Y? That's impossible since X and Y intersect but all from X and all from Y would mean all from intersection {B,C}, but size 3 >2. So 20-4-4=12. That gives 12 sets. But we also have sets that include some from {G,H,I,J}? Those would have at least one from X and one from Y, but then the third could be from outside. However, if we include an outside person, we still need at least one from X and one from Y among the other two. So there are more than 12? Let's recalc: The formula gave 12, but that seems low. Let's check: Total 120. No X: 56, No Y: 56, both: 4. So 120-56-56+4=12. That means only 12 sets have at least one from X and one from Y. But intuitively, if we pick any set that includes, say, Alice (X) and Evan (Y), and then any third person, that should be many. For example, {A,E,G} has A from X, E from Y, so it should count. But according to formula, it does. Let's test: {A,E,G} is in the 12? Let's list all 12 by enumerating from X∪Y only? Wait, the formula counts all sets from the 10 employees. Let's manually count sets with at least one from X and one from Y. We can do complementary counting: total sets with no X or no Y. No X: choose 3 from 8 (not X). That includes sets that might have Y or not. No Y: choose 3 from 8 (not Y). Intersection: choose 3 from 4 (neither). So 120 - 56 - 56 + 4 = 12. That is correct. So there are only 12 such sets. But wait, that means most sets do not have both an X and a Y member. For example, a set with two from X and one from outside (not Y) would have no Y, so it's excluded. A set with one from X, one from Y, and one from outside should be counted. Let's count those: choose 1 from X (4 ways), 1 from Y (4 ways), and 1 from outside X∪Y (4 ways: G,H,I,J). That gives 444=64 sets. But these sets have exactly one from X, one from Y, and one outside. They should be included. But our formula gave only 12, which is a contradiction. So my calculation of no X and no Y must be wrong. Let's recompute no X: employees not in X: {E,F,G,H,I,J}? Wait, X={A,B,C,D}. So not in X: {E,F,G,H,I,J}? That's 6? But earlier I said 8. Let's list all 10 employees: A,B,C,D,E,F,G,H,I,J. Not in X: E,F,G,H,I,J = 6 employees. So C(6,3)=20. Not in Y: Y={B,C,E,F}. So not in Y: A,D,G,H,I,J = 6 employees. C(6,3)=20. Not in X and not in Y: employees not in X and not in Y: that is {G,H,I,J}? Also check: E,F are in Y, so they are in Y, so they are not in "not Y". So intersection is {G,H,I,J} = 4. C(4,3)=4. Then total = 120 - 20 - 20 + 4 = 84. That makes more sense. So there are 84 sets with at least one from X and one from Y. Now we need to exclude those containing a forbidden pair. Forbidden pairs: (A,D), (B,C), (C,E). We need to count how many of the 84 sets contain at least one of these pairs. We can use inclusion-exclusion on the three pairs. Let S1 = sets containing (A,D). S2 = sets containing (B,C). S3 = sets containing (C,E). We want |S1 ∪ S2 ∪ S3|. First, count each: For a given pair, we need to choose a third person from the remaining 8 employees (since 10 total, 2 in pair). But we must ensure the set still has at least one from X and one from Y. However, if the pair already includes members from X and Y, the condition may be satisfied automatically. Let's check each pair: (A,D): A in X, D in X. So both from X. To have at least one from Y, the third person must be from Y. Y = {B,C,E,F}. So third person must be one of B,C,E,F. That gives 4 sets. But also need to ensure no other forbidden pair? We'll handle overlaps later. So |S1| = 4. (B,C): both in X and Y (since B,C are in both). So any third person will still have at least one from X and one from Y (since B and C are in both). So third person can be any of the remaining 8 employees. So |S2| = 8. (C,E): C in X and Y, E in Y. So we have at least one from X (C) and at least one from Y (C and E). So any third person works. So |S3| = 8. Now intersections: S1 ∩ S2: sets containing (A,D) and (B,C). That's 4 people: A,D,B,C. To form a 3-person set, we can't have both pairs because that would require 4 people. So no set can contain both (A,D) and (B,C) since they are disjoint? Actually (A,D) and (B,C) are disjoint pairs, so a 3-person set cannot contain both. So intersection empty. Similarly, S1 ∩ S3: (A,D) and (C,E) are disjoint? A,D,C,E are 4 distinct, so cannot both be in a 3-person set. So empty. S2 ∩ S3: (B,C) and (C,E) share C. So a set containing both pairs would have B,C,E. That's 3 people. So that set is {B,C,E}. Does it satisfy at least one from X and one from Y? Yes. So there is 1 such set. S1 ∩ S2 ∩ S3: would require A,D,B,C,E which is 5 people, impossible. So by inclusion-exclusion, number of sets with at least one forbidden pair = |S1|+|S2|+|S3| - |S1∩S2| - |S1∩S3| - |S2∩S3| + |S1∩S2∩S3| = 4+8+8 - 0 - 0 - 1 + 0 = 19. So valid sets = 84 - 19 = 65. But wait, we must also ensure that the set does not contain any other forbidden pair? We only have three forbidden pairs. So 65 seems the answer. However, we need to double-check if any set counted in S1, S2, S3 might not satisfy the X and Y condition? We already ensured that for each pair, the third person was chosen to satisfy the condition. For S1, we required third from Y, so condition met. For S2 and S3, condition automatically met. So all sets in S1, S2, S3 satisfy the condition. So the subtraction is valid. Thus 65 valid task forces. But let's verify with a smaller example or check for overcounting? The inclusion-exclusion seems correct. So answer: 65.