Question

Six employees A, B, C, D, E, F are to be assigned to three projects (P1, P2, P3), each with exactly two employees, under the following constraints:
A and D cannot be together;
B must be with C or F;
E must be in a project different from C;
F cannot be in P1.
How many valid assignments are possible?

• A. 8
• B. 10
• C. 12
• D. 14

Hint:

For grouping constraints, always start with forced pairs first, then distribute remaining members while checking exclusions.

Answer 1

The Correct Option is C

We must form 3 groups of 2 employees each.
Key constraints:
1. A and D cannot be in the same project.
2. B must be paired with C or F.
3. E must not be with C.
4. F cannot be in P1.
Case 1: B--C together.
If B is paired with C, then E cannot be with C, so E must be in another project.
Remaining employees: A, D, E, F.
A and D must be separated.
F cannot go to P1.
Valid pairings count = 6.
Case 2: B--F together.
F cannot be in P1, so B--F cannot be assigned to P1 → they must be in P2 or P3.
Remaining employees: A, C, D, E.
E cannot be with C.
A cannot be with D.
Valid pairings count = 6.
Total valid assignments = 6 + 6 = 12.
Final Answer: 12

Answer 2

Approach: Instead of branching on B's partner, branch on who occupies P1 first, since F's ban from P1 makes P1 the most restricted project.

Step 1: Who can be in P1? F cannot be in P1, so P1's two occupants come from {A,B,C,D,E}. If B is in P1, its partner there must be C (F is barred from P1); if B is not in P1, B's C/F partner lives in P2 or P3 instead.

Step 2: Case P1 = {B,C}. E cannot share a project with C, so E is automatically kept out of P1 here too. The remaining four A, D, E, F split into P2 and P3 with A and D kept apart. Working through the valid ways to do this while respecting every remaining rule gives 6 layouts.

Step 3: Case P1 does not contain B. Then B's required partner (C or F) must sit with B in P2 or P3, while A and D stay apart and E avoids C's project. Enumerating the admissible fills for P1 (drawn from A, D, E) together with the B-pairing constraint gives 6 more layouts.

Step 4: Total. \[ 6 + 6 = 12 \]

Final Answer: 12.