UPSC Prelims 2009 Q98: Science & Technology – Quantitative Aptitude

UPSC Prelims 2009, General Studies Paper I · Science & Technology

How many three - digit numbers can be generated from 1, 2, 3, 4, 5, 6, 7, 8, 9 such that the digits are in ascending order?

Answer: (d) 84

Explanation

Counting three-digit numbers with strictly ascending digits

Using digits 1 through 9 (no zero, no repetition), we need three-digit numbers where each digit is strictly greater than the one before it.

Step	Reasoning
Key insight	Any choice of 3 distinct digits from 1-9 can be arranged in exactly ONE way to be strictly ascending
Count ways to choose 3 digits from 9	This is a simple combination: C(9,3)
Calculate	C(9,3) = 9!/(3!×6!) = 84

Why it's a combination, not a permutation:

Point	Detail
Any 3 chosen digits	Have exactly one valid ascending arrangement
So the count of numbers	Equals the count of ways to choose the 3 digits, i.e., C(9,3)

Answer: (d) 84.

(a) Incorrect. 80 does not match the exact C(9,3) calculation.
(b) Incorrect. 81 is close but not the precise combinatorial result.
(c) Incorrect. 83 also does not match the exact calculation.
(d) CORRECT. C(9,3) = 84, the exact number of ways to choose 3 digits (each choice giving exactly one ascending arrangement).

Any set of 3 distinct digits from 1-9 has exactly one strictly ascending arrangement, so the count of such 3-digit numbers equals C(9,3) = 84.

Three-digit ascending-digit numbers from 1-9: C(9,3) = 84.

Treating this as a permutation problem (ordering matters) instead of recognising only ONE valid order exists per digit set.
Making an arithmetic slip while computing the combination C(9,3).
Forgetting that digits cannot repeat in a strictly ascending sequence.

'Strictly ascending digits' problems: just count the ways to CHOOSE the digits (combination) — each choice has exactly one valid arrangement.

NCERT Class XI Mathematics — Permutations and Combinations
R. S. Aggarwal — Quantitative Aptitude for Competitive Examinations (Permutations and Combinations)