Question

E-1, E-2 and E-3 are three engineering students writing their assignments at night. Each of them starts at a different time and completes at a different time. The digit in their name and the order of their starting and completing the assignment is certainly not the same. The last student to start is the first to complete the assignment.
Who is the last student to complete the assignment?

• A. E-1
• B. E-2
• C. E-3
• D. Cannot be decided

Hint:

First work out the starting order and who finishes first, then the last two completion slots follow by elimination.

Answer 1

The Correct Option is A

Step 1: Recall the setup.
Let S(Ei) be the starting-order position and C(Ei) be the completion-order position of student E-i, with S(Ei) not equal to i and C(Ei) not equal to i for each student, and the last student to start being the first to complete.
Step 2: Recall the valid starting order.
As established from the derangement analysis, the only valid starting order is E-3 first (S = 1), E-1 second (S = 2), E-2 last (S = 3). The alternative arrangement forces E-1 to both start last and complete first, which breaks the rule that C(E1) cannot equal 1, so it is ruled out.
Step 3: Apply the last-to-start-is-first-to-complete rule.
Since E-2 starts last, E-2 must be first to complete, so C(E2) = 1.
Step 4: Assign the remaining completion positions.
Positions 2 and 3 remain for E-1 and E-3. Since C(E3) cannot equal 3 (E-3's own digit), E-3 must take position 2, so C(E3) = 2, leaving C(E1) = 3.
Step 5: Check this is a valid derangement.
C(E1) = 3 (not 1), C(E2) = 1 (not 2), C(E3) = 2 (not 3): none of the completion positions match the student's own digit, so this is consistent.
Step 6: Answer the question.
Since C(E1) = 3, E-1 is the last student to complete the assignment, which is option (1).