UPSC NDA II 2015 Mathematics Paper

Concept:The product of three digits is odd only when all three digits are odd. The problem involves selecting 3 digits from 1 to 9 without repetition. Use combinations because the order of selection does not matter.Explanation:Total digits available: 1, 2, 3, 4, 5, 6, 7, 8, 9 (9 digits). Odd digits: 1, 3, 5, 7, 9 (5 odd digits). Step 1: Total number of ways to choose any 3 digits from 9 is $^{9}C_{3} = \frac{9!}{3!6!} = 84$. Step 2: Number of ways to choose 3 digits all odd from the 5 odd digits is $^{5}C_{3} = \frac{5!}{3!2!} = 10$. Step 3: Probability = (favorable outcomes) / (total outcomes) = $\frac{10}{84} = \frac{5}{42}$. Alternatively, if order mattered (forming 3-digit numbers), total arrangements = $9 \times 8 \times 7 = 504$, and favorable arrangements = $5 \times 4 \times 3 = 60$, giving the same probability $\frac{60}{504} = \frac{5}{42}$.Answer:$\frac{5}{42}$ Thus, option C is correct.