UPSC Prelims 2010 Q87: Science & Technology – Quantitative Aptitude
Each person's performance compared with all other persons is to be done to rank them
subjectively. How many comparisons are needed in total, if there are 11 persons?
- (a) 66
- (b) 55
- (c) 54
- (d) 45
Answer: (b) 55
Explanation
Counting pairwise comparisons among 11 persons
Ranking each person against every other person once requires counting all unique pairs.
| Step | Calculation |
|---|---|
| Formula | n(n−1)/2, where n = number of persons |
| Substitute n = 11 | 11 × 10 / 2 |
| Compute | 110 / 2 = 55 |
Why this formula applies:
| Point | Detail |
|---|---|
| Each person compared to | 10 others |
| Total person-comparisons | 11 × 10 = 110 (counted twice, once per person in the pair) |
| Unique comparisons | Divide by 2: 55 |
Answer: (b) 55.
Why each option is right or wrong
- (a) Incorrect. 66 corresponds to n=12 (12×11/2=66), not 11 persons.
- (b) CORRECT. Using n(n−1)/2 with n=11 gives 11×10/2 = 55 unique comparisons.
- (c) Incorrect. 54 does not match the correct formula's result for 11 persons.
- (d) Incorrect. 45 corresponds to n=10 (10×9/2=45), not 11 persons.
Key point
For pairwise comparison among n persons, total unique comparisons = n(n−1)/2. With n=11, that's 11×10/2 = 55.
Remember
11 persons, pairwise comparisons: 11×10/2 = 55.
Common mistakes
- Forgetting to divide by 2 after multiplying n by (n−1), double-counting each comparison.
- Using the wrong value of n (e.g., 10 or 12) instead of the given 11 persons.
- Confusing this with a different combinatorial counting problem (like permutations instead of combinations).
Exam tip
Pairwise comparison count formula: n(n−1)/2 — identical in structure to the round-robin tournament match-count formula.
Sources
- NCERT Class XI Mathematics — Permutations and Combinations
- R. S. Aggarwal — Quantitative Aptitude for Competitive Examinations (Permutations and Combinations)