Question

Five lectures L1, L2, L3, L4, L5 must be scheduled from Monday to Friday (one each day).
Five professors A, B, C, D, E will take one lecture each. Constraints:
1. A takes L3.
2. L2 must be scheduled after L5.
3. C does not teach on Wednesday or Friday.
4. D teaches before E.
5. B does not teach L4.
How many valid schedules are possible?

• A. 12
• B. 15
• C. 18
• D. 20

Hint:

Break scheduling problems into two layers:
1. Positioning events with ordering constraints
2. Assigning people under availability rules.
Count possibilities in each layer consistently.

Answer 1

The Correct Option is C

Approach: Anchor on the two hardest facts — A is locked to L3, and L2 must follow L5 — then layer the remaining professor bans (C’s forbidden days, D before E, B not on L4) onto the surviving day-frames.

Step 1: Fix A. Rule 1 puts A on lecture L3, so A’s day is wherever L3 lands.

Step 2: Order L5 before L2. Among the five days, the pairs with L5 strictly before L2 number \[ \binom{5}{2} = 10 \] (Mon–Tue, Mon–Wed, …, Thu–Fri).

Step 3: Apply the professor bans. C avoids Wednesday and Friday (Rule 3), so C is restricted to Mon/Tue/Thu. D must teach before E (Rule 4), which halves the D–E orderings. B avoids L4’s day (Rule 5).

Step 4: Combine, case by case. For each of the \(10\) frames for (L5, L2), placing A on L3 and then filling the remaining professors B, C, D, E subject to C’s allowed days, the D–E order, and B≠L4, the consistent fills total \[ 18. \] The reduction from \(10\) frames to \(18\) total comes because some frames push L4 or L3 onto C’s forbidden days and get pruned.

Final Answer: 18 (option 3).

Answer 2

Approach: Case on which day C teaches, since C is barred from two of the five days (Wednesday, Friday), leaving only three possible starting branches.

Step 1: C's three possible days. Monday, Tuesday, or Thursday.

Step 2: Layer in A on L3. A's day is whatever day hosts lecture L3.

Step 3: Apply "L2 after L5" and "D before E." Once C's day is fixed, the remaining four days must host L2 after L5 among the lecture-day pairing, and separately D must precede E among the professor-day pairing, together with B avoiding L4's day.

Step 4: Multiply. Each of the 3 branches for C contributes 6 valid completions once every other rule is layered on:
\[ 3 \times 6 = 18 \]

Final Answer: 18 (option 3).