UPSC NDA I 2024 Mathematics Paper

Concept:Probability is the ratio of favorable outcomes to total outcomes when selecting three distinct numbers from 1 to 15.Explanation:Total number of ways to choose 3 distinct numbers from the first 15 natural numbers is $\binom{15}{3} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 455$.We need sets $\{a,b,c\}$ with $a \times b = c$, all numbers distinct and between 1 and 15.Since the numbers are unordered, consider $a < b$.The product must be at most 15 and different from both $a$ and $b$.List all such ordered pairs $(a,b)$ with $a < b$ and $a \times b \le 15$:$(2,3) \to 6$, $(2,4) \to 8$, $(2,5) \to 10$, $(2,6) \to 12$, $(2,7) \to 14$, $(3,4) \to 12$, $(3,5) \to 15$.Pairs with $a=1$ give $1 \times b = b$, so the third number equals $b$, which violates distinctness. Hence they are invalid.Thus, there are exactly 7 favorable sets.So probability = $\frac{7}{455} = \frac{1}{65}$.Answer:$\frac{1}{65}$