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 first student to start writing the assignment?

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

Hint:

Use the fact that no student's start or finish position can match their own name digit, and that the last starter finishes first, to build a valid order.

Answer 1

The Correct Option is C

Step 1: Set up the notation.
Let S(Ei) be the starting-order position (1 = first to start, 3 = last to start) of student E-i, and C(Ei) be the completion-order position (1 = first to complete, 3 = last to complete). Both S and C are permutations of {1, 2, 3}. The clue tells us that for every student, the name digit i is different from both S(Ei) and C(Ei), so S(Ei) is not equal to i, and C(Ei) is not equal to i, for i = 1, 2, 3.
Step 2: List the possible starting-order arrangements.
A permutation of {1, 2, 3} where no position matches the student's own name digit is called a derangement. For 3 items, there are exactly 2 such derangements: Arrangement A, with S(E1) = 2, S(E2) = 3, S(E3) = 1, and Arrangement B, with S(E1) = 3, S(E2) = 1, S(E3) = 2.
Step 3: Apply the rule that the last student to start is the first to complete.
In Arrangement A, the last to start (S = 3) is E-2, so E-2 must be first to complete, meaning C(E2) = 1.
In Arrangement B, the last to start (S = 3) is E-1, so E-1 must be first to complete, meaning C(E1) = 1. But this directly breaks the rule that C(Ei) cannot equal i, since E-1's own digit is 1. So Arrangement B is impossible.
Step 4: Conclude the valid starting order.
Only Arrangement A works: E-3 starts first (S = 1), E-1 starts second (S = 2), E-2 starts last (S = 3).
Step 5: Cross-check with the completion order.
We know C(E2) = 1. The remaining positions 2 and 3 go to E-1 and E-3, and since C(E3) cannot equal 3, we get C(E3) = 2 and C(E1) = 3. This is a valid derangement, so everything is fully consistent.
Step 6: Answer the question.
Since E-3 has S(E3) = 1, E-3 is the first student to start writing the assignment, which is option (3).