Question

Six employees — A, B, C, D, E, F — each specialize in exactly one of three skills: Data, Design, or Marketing (two per skill).
1. A and D do not share a skill.
2. B’s skill is the same as either E or F (but not both).
3. C is not in Marketing.
How many valid assignments of skills are possible?

Hint:

Whenever a constraint links three people (like “B matches exactly one of E or F”), handle skill capacities first, then apply the exclusive condition—it always halves the possibilities.

Answer 1

Correct Answer: 16

There are 6 employees and 3 skills, with exactly two people per skill. Let the skills be: \[ D_a = \text{Data},\quad D_e = \text{Design},\quad M = \text{Marketing} \] Step 1: Assign C’s skill. 
C is not in Marketing. So C is either in Data or Design. 
We must consider both possibilities. 
Case 1: C is in Data. 
Then Data has one slot remaining. 
Let Data = C, X. 
A and D cannot share a skill, so they cannot both be Data or both be Design or both be Marketing.
We enumerate possible placements for A and D while respecting the “two per skill” capacity. 
We find that valid A–D pairs are: \[ (C, A)\text{ in Data and }D\text{ in Design or Marketing}, \] \[ (C, D)\text{ in Data and }A\text{ in Design or Marketing} \] This yields exactly 4 valid A–D distributions after checking capacity limits. 
Next, handle B’s condition. 
Step 1A: B must match exactly one of E or F. 
So, the pair (B,E,F) must have exactly one match: \[ B=E\neq F \quad\text{or}\quad B=F\neq E \] Given remaining slots per skill from the A–D–C placements, each A–D layout yields 2 valid assignments for (B,E,F). 
Thus, Case 1 gives: \[ 4 \times 2 = 8 \text{ valid assignments.} \] Case 2: C is in Design. 
By symmetry with Case 1 (Data ↔ Design), all calculations mirror perfectly. 
Thus Case 2 also gives: \[ 8 \text{ valid assignments.} \] Step 2: Combine both cases. 
\[ 8 + 8 = 16 \] Final Answer: \(\boxed{16}\)