Title: CAT 24 Slot 3 DILR Solution (Indian Exam Answer)
Question Context (Hypothetical Example for Clarity):
In a logistics company, five cities (P, Q, R, S, T) are connected by road, air, or rail. The following conditions apply:
Each city is connected to every other city by exactly one mode.
Road connections are always even in number.
Air and rail connections are never between consecutive cities (alphabetically).
City R has the most rail connections.
How many valid configurations are possible?
Solution Steps
Understand the Problem Structure:
Total cities: 5 ⇒ Total possible connections: C(5,2) = 10.
Each connection is either Road (R), Air (A), or Rail (L).
Constraints:
Road connections must be even (0, 2, 4, 6, 8, 10).
Air/Rail connections ≠ between consecutive cities (P-Q, Q-R, R-S, S-T).
Analyze Constraints:
Consecutive City Pairs: 4 pairs (P-Q, Q-R, R-S, S-T). These cannot be Air or Rail.
⇒ These 4 pairs must be Road.
Road Count: Since 4 road connections are fixed (consecutive pairs), the total road count must be even.
⇒ Remaining roads: 10 total - 4 fixed = 6. We need to add 0, 2, 4, or 6 more roads.
Rail Connections: R must have the most rail links.
Deduce Remaining Connections:
Non-consecutive pairs: 10 total - 4 consecutive = 6 pairs.
These 6 pairs can be Air or Rail, but Rail must dominate at city R.
Focus on City R:
City R is connected to P, Q, S, T.
Consecutive pairs involving R: Q-R and R-S (already Road).
Non-consecutive pairs: P-R, R-T.
To maximize Rail at R, assign Rail to P-R and R-T.
⇒ Rail connections for R: 2 (P-R, R-T).
Assign Remaining Non-Consecutive Pairs:
Remaining non-consecutive pairs: P-Q (Road), Q-S, Q-T, P-S, P-T, S-T (Road).
Non-consecutive pairs to assign: Q-S, Q-T, P-S, P-T.
These must be Air or Rail, but Rail cannot exceed Air in these pairs (to avoid violating Road parity).
Calculate Valid Configurations:
Total non-consecutive pairs to assign: 4 (Q-S, Q-T, P-S, P-T).
Each can be Air or Rail, but total Rail connections must not exceed Air to maintain Road parity.
Possible Rail counts here: 0, 2 (since 4 is even but would make Rail dominant).
Case 1: 0 Rail ⇒ All Air. Valid.
Case 2: 2 Rail ⇒ Choose 2 pairs for Rail, others Air. C(4,2) = 6 ways.
Total configurations: 1 + 6 = 7.
Answer:
There are 7 valid configurations satisfying all constraints.
Key Takeaways for CAT DILR:
Prioritize constraints (e.g., consecutive city pairs).
Break problems into smaller subproblems (e.g., focus on city R first).
Use combinatorial logic (C(n,k)) for permutations.
Verify parity (even/odd) and dominance conditions.
Let me know if you need further clarification!
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