Question

Which employee is always in the team with E5?

• A. E1
• B. E2
• C. E3
• D. None
• A. 3
• B. 4
• C. 5
• D. 6
• A. E2
• B. E3
• C. E4
• D. E5

Answer 1

The Correct Option is D

We need to find if any employee is always in the team with E5.
- Step 1: Apply condition 3. E5 is always selecte(d) Team: [E5, \_, \_].
- Step 2: Select 2 more from E1, E2, E3, E4. Conditions:
- If E1, then not E2 (condition 1).
- Not both E3 and E4 (condition 2).
- Step 3: List possible teams.
- E1, E5, and (E3 or E4, not both):
- E1, E3, E5.
- E1, E4, E5.
- E2, E5, and (E3 or E4, not both):
- E2, E3, E5.
- E2, E4, E5.
- Step 4: Check for always include(d)
- E1: Not in teams with E2 (e.g., E2, E3, E5).
- E2: Not in teams with E1 (e.g., E1, E3, E5).
- E3: Not in teams with E4 (e.g., E1, E4, E5).
- E4: Not in teams with E3 (e.g., E1, E3, E5).
- Step 5: Conclusion. No employee is in every team with E5.
- Step 6: Check options.
- (a) E1: Incorrect, not in E2 teams.
- (b) E2: Incorrect, not in E1 teams.
- (c) E3: Incorrect, not in E4 teams.
- (d) None: Correct.
Thus, the answer is d.

Answer 2

We need the number of valid teams of 3 employees.
- Step 1: Apply conditions. E5 always selecte(d) Team: [E5, \_, \_].
- Step 2: Select 2 from E1, E2, E3, E4. Conditions:
- E1 and E2 not together.
- E3 and E4 not together.
- Step 3: List combinations.
- With E1 (not E2): E1, E3, E5; E1, E4, E5.
- With E2 (not E1): E2, E3, E5; E2, E4, E5.
- Step 4: Count teams.
- E1, E3, E5.
- E1, E4, E5.
- E2, E3, E5.
- E2, E4, E5.
- Total = 4.
- Step 5: Verify. All teams satisfy conditions 1 and 2.
- Step 6: Check options.
- (a) 3: Incorrect.
- (b) 4: Correct.
- (c) 5: Incorrect.
- (d) 6: Incorrect.
Thus, the answer is b.

Answer 3

We need the employee who cannot be in the team if E1 is selecte(d)
- Step 1: Apply conditions. E5 always selecte(d) E1 selected: Team [E1, E5, \_].
- Step 2: Check condition 1. If E1, then not E2.
- Step 3: Check others.
- E3 or E4 can be selected (not both, condition 2).
- E5 is already selecte(d)
- Step 4: Conclusion. E2 cannot be selecte(d)
- Step 5: Check options.
- (a) E2: Correct.
- (b) E3: Incorrect, E3 possible (E1, E3, E5).
- (c) E4: Incorrect, E4 possible (E1, E4, E5).
- (d) E5: Incorrect, E5 always selecte(d)
Thus, the answer is a.

Answer 4

We need the number of valid teams with E3.
- Step 1: Apply conditions. E5 always selected, E3 selected: Team [E3, E5, \_].
- Step 2: Select third employee.
- E2 and E4 possible (E4 not with E3, condition 2).
- E1 not with E2 (condition 1).
- Step 3: List teams.
- E3, E5, E1 (E2 excluded).
- E3, E5, E2 (E1 excluded).
- Step 4: Count. 2 teams: [E1, E3, E5], [E2, E3, E5].
- Step 5: Verify. Both satisfy all conditions.
Thus, the answer is 2.