Title: CAT 2017 QA Slot 2 Solution (Indian Game Answer)
Problem Statement (Hypothetical Example):
Five players (A, B, C, D, E) participate in a game. Each player rolls a die once. The game’s rule: If a player rolls a number higher than the previous player, they win a point. The first player starts. In how many ways can Player A win exactly 2 points?
Solution:
Understanding the Game Mechanics:
Points are awarded if a player rolls a higher number than the immediately preceding player.
Player A starts, so they cannot earn a point from a prior player.
Player A can win points from Players B, C, D, or E if their roll is higher than the respective predecessor.
Objective:
Count sequences where Player A (P1) has exactly 2 instances where their roll > the previous player’s roll.
Approach:
Use permutations with constraints.
Focus on A’s position (P1) and track where P1 > P_{n-1}.
Step-by-Step Calculation:
Total Possible Outcomes: (6^5 = 7776).
Constraints: A must have exactly 2 wins.
Win Cases for A:
A can win against B, C, D, or E.
Choose 2 out of these 4 positions where A > predecessor.
Use combinations: (\binom{4}{2} = 6).
Probability Calculation:
For each chosen position (e.g., A > B and A > C), calculate valid permutations:
A’s roll must be > predecessor’s roll.
Example: If A > B and A > C:
A: 2–6 (5 choices).
B: 1–(A–1).
C: 1–(A–1).
D and E: Any roll (6 choices each).
Total for this case: (5 \times 1 \times 1 \times 6 \times 6 = 180).
Multiply by 6 (from combinations): (6 \times 180 = 1080).
Final Answer:
The number of ways Player A can win exactly 2 points is 1080.
Key Takeaways:
Break down the problem into manageable constraints.
Use combinatorial logic to count valid sequences.
Prioritize player positions where A can influence outcomes.
Note: This is a hypothetical example. For the actual CAT 2017 QA Slot 2 question, refer to official resources or prep materials like Target CAT 2017 by Arun Shukla.
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