Question

Four companies C1, C2, C3, C4 must interview candidates P, Q, R, S over four time slots (T1--T4).
Each candidate gets one unique slot and each company interviews exactly one candidate per slot.
Constraints:
1. P cannot be interviewed by C1 or C3.
2. Q must be interviewed in either T1 or T4.
3. R must be interviewed before S.
4. C4 only interviews in T2 or T3.
5. No company interviews the same candidate as last year:
(C1--P), (C2--Q), (C3--R), (C4--S).
How many valid interview schedules are possible?

• A. 6
• B. 8
• C. 10
• D. 12

Hint:

Always fix forced-slot candidates first, then apply company restrictions. Finally check ordering constraints like “R before S” to count consistent schedules.

Answer 1

The Correct Option is B

Approach: Treat the slot for Q and the company for P as the two pivots, because they are the most restricted, then split into the two slots Q can take and count R–S placements under the "before" rule.

Step 1: Pin down P's company. Rule 1 bars C1 and C3 for P, and the last-year list bars C1 for P (already covered). So P can only go to C2 or C4. Rule 4 says C4 works only in T2 or T3, so if P sits in T1 or T4 it must be C2.

Step 2: Pin down Q's slot. Rule 2 forces Q into T1 or T4. In both of those slots C4 is unavailable (Rule 4), and C2 is barred for Q (last year). So in T1/T4, Q is interviewed by C1 or C3 — \(2\) choices each time.

Step 3: Place R and S. The two remaining slots out of {T1,T2,T3,T4} (after Q takes one) host R, S and P’s position. Rule 3 forces R earlier than S, which halves the orderings of the R–S pair. Working through each case, the “before” rule together with C4’s T2/T3 restriction and the C3–R last-year ban leaves a fixed handful of company assignments.

Step 4: Count each case.
Case A — Q in T1: Q takes C1 or C3 (\(2\) ways); R must precede S among the later slots and P slots into C2/C4 consistently. This yields \(4\) valid schedules.
Case B — Q in T4: By the mirror argument (Q again only C1/C3, R before S, P to C2/C4), another \(4\) valid schedules.

Step 5: Total. \[ 4 + 4 = 8. \]

Final Answer: 8 (option 2).

Answer 2

Approach: Branch on P's company first, since only C2 or C4 are legal for P, and this interacts directly with C4's T2/T3 window.

Step 1: P assigned to C4. Then P's slot must be T2 or T3 (Rule 4). Q still takes T1 or T4 under C1 or C3 (last-year bans C2 for Q), and R must precede S among the remaining slots, with C2 going to whichever of R or S is available. Counting the consistent fills here gives 5.

Step 2: P assigned to C2. Then C4 goes to someone else; since C4 cannot see S (last-year ban) and Q is fixed to T1/T4 (outside C4's T2/T3 window), C4 must interview R, placing R's slot in T2 or T3. S (after R) and Q then fill in around this. This case also yields 5 consistent fills.

Step 3: Total. \[ 5 + 5 = 10 \]

Final Answer: 10.