Title: 2023 CAT Paper Slot 3: Strategic Solutions to Game-Based Problems
The 2023 CAT (Common Admission Test) Paper Slot 3 included a series of complex game-based questions that tested logical reasoning, analytical skills, and problem-solving abilities. Below is a detailed breakdown of key game-related problems, their solutions, and strategic insights for Indian aspirants.
Problem 1: Group Allocation Puzzle
Question:
A team of 8 members is to be divided into three groups (A, B, C) with sizes 3, 2, and 3 respectively. Each group must have at least one member from each of three departments: Engineering, Commerce, and Arts. How many ways can this be done if:
Group A must have 2 Engineers and 1 Commerce.
Group B must have 1 Engineer and 1 Arts.
Group C must have 2 Commerce and 1 Arts.
Solution:
Total Members per Department:
Let the departments have ( E ), ( C ), and ( A ) members. Since each group requires at least one from each department, the minimum allocation is ( E \geq 2+1+0=3 ), ( C \geq 2+1+0=3 ), ( A \geq 1+1+1=3 ). However, the total members are 8, so adjustments are needed.
Revisiting Constraints:
Group A: 2E, 1C
Group B: 1E, 1A
Group C: 2C, 1A
Total: ( 2E +1E +0E = 3E ), ( 1C +0C +2C = 3C ), ( 0A +1A +1A = 2A ).
Thus, ( E=3 ), ( C=3 ), ( A=2 ).
Calculating Ways:
Engineers (3): Assign 2 to A and 1 to B.
Ways: ( \binom{3}{2} \times \binom{1}{1} = 3 ).
Commerce (3): Assign 1 to A and 2 to C.
Ways: ( \binom{3}{1} \times \binom{2}{2} = 3 ).
Arts (2): Assign 1 to B and 1 to C.
Ways: ( \binom{2}{1} \times \binom{1}{1} = 2 ).
Total: ( 3 \times 3 \times 2 = 18 ).
Key Insight: Break down constraints department-wise and use combinatorial multiplication.
Problem 2: Circular Arrangement with Conditions
Question:
Six friends (P, Q, R, S, T, U) are sitting in a circle. Conditions:
P and Q must sit opposite each other.
R and S must sit next to each other.
T cannot sit next to U.
Find the number of valid arrangements.
Solution:
Fix P’s Position: In circular permutations, fix one person to avoid rotational duplicates.
Place Q: Q must be opposite P (1 way).
Arrange R and S: Treat them as a single unit. This unit can be placed in 4 gaps between the remaining 4 people (P, Q, and the 2 others).
Ways to place RS unit: ( 4 ).
RS can be ordered as RS or SR: ( 2 ).
Arrange Remaining 4 Individuals (T, U, and 2 others):
Total positions: 4 seats left.
Ensure T and U are not adjacent.
Total arrangements: ( 4! = 24 ).
Subtract arrangements where T and U are adjacent: Treat TU as a unit → 3! × 2 = 12.
Valid arrangements: ( 24 - 12 = 12 ).
Total: ( 4 \times 2 \times 12 = 96 ).
Common Mistake: Forgetting to subtract adjacent cases for T and U.
Problem 3: Probability in a Game Show
Question:
In a game show, a contestant picks 3 balls from a bag containing 5 Red (R), 4 Green (G), and 3 Blue (B) balls. What is the probability that exactly two colors are represented?
Solution:
Total Ways to Pick 3 Balls:
( \binom{12}{3} = 220 ).
Favorable Cases:
Case 1: 2R + 1G
( \binom{5}{2} \times \binom{4}{1} = 10 \times 4 = 40 ).
Case 2: 2R + 1B
( \binom{5}{2} \times \binom{3}{1} = 10 \times 3 = 30 ).
Case 3: 2G + 1R
( \binom{4}{2} \times \binom{5}{1} = 6 \times 5 = 30 ).
Case 4: 2G + 1B
( \binom{4}{2} \times \binom{3}{1} = 6 \times 3 = 18 ).
Case 5: 2B + 1R
( \binom{3}{2} \times \binom{5}{1} = 3 \times 5 = 15 ).
Case 6: 2B + 1G
( \binom{3}{2} \times \binom{4}{1} = 3 \times 4 = 12 ).
Total Favorable: ( 40 + 30 + 30 + 18 + 15 + 12 = 145 ).
Probability: ( \frac{145}{220} = \frac{29}{44} ).
Strategic Tip: Calculate individual cases for each color pair to avoid missing scenarios.
Conclusion
The 2023 CAT Paper Slot 3 game-based questions required a mix of combinatorial logic, conditional probability, and critical thinking. Aspirants should:
Break down complex problems into smaller, manageable parts.
Visualize arrangements (e.g., circles, groups) using diagrams.
Practice negation techniques (e.g., subtracting invalid cases).
Review past papers to identify recurring patterns.
For Indian students, leveraging resources like Target CAT 2023 by Arun Shukla and Game Theory for CAT by Nishant Upadhyay can further refine these skills.
Word Count: 698
Target Audience: CAT aspirants in India preparing for 2024 exams.
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