UPSC Prelims 2007 Q137: Science & Technology – Quantitative Aptitude
Five balls of different colours are to be placed in three different boxes such that any box
contains at least one ball. What is the maximum number of different ways in which this can be
done?
- (a) 90
- (b) 120
- (c) 150
- (d) 180
Answer: (c) 150
Explanation
5 different balls into 3 different boxes, no box empty
This counts onto (surjective) functions from 5 balls to 3 boxes.
Use inclusion-exclusion:
| Step | Calculation | Value |
|---|---|---|
| All ways (any box may be empty) | 3⁵ | 243 |
| Remove ways with at least one box empty | − C(3,1) × 2⁵ | − 96 |
| Add back ways with two boxes empty | + C(3,2) × 1⁵ | + 3 |
| Total | 243 − 96 + 3 | 150 |
Check by splits: (3,1,1) → 60 ways; (2,2,1) → 90 ways; 60 + 90 = 150.
Answer: (c) 150.
Why each option is right or wrong
- (a) Incorrect. 90 counts only the (2,2,1) split.
- (b) Incorrect. 120 = 5! — that would be arranging 5 balls in a line.
- (c) CORRECT. 3⁵ − 3·2⁵ + 3 = 150.
- (d) Incorrect. 180 over-counts — it double counts some splits.
Key point
Distinct balls into distinct boxes, none empty: 3⁵ − 3·2⁵ + 3·1⁵ = 150.
Remember
5 distinct balls, 3 boxes, none empty = 150.
Common mistakes
- Using 3⁵ = 243 and forgetting the 'no box empty' rule.
- Subtracting 3·2⁵ but forgetting to add back the 3 cases with two empty boxes (→ 147).
- Counting only one split pattern (3,1,1) or (2,2,1).
Exam tip
Inclusion-exclusion pattern: kⁿ − C(k,1)(k−1)ⁿ + C(k,2)(k−2)ⁿ − …
Sources
- NCERT Class XI — Mathematics, Ch 6 (Permutations and Combinations)
- R. S. Aggarwal — Quantitative Aptitude for Competitive Examinations (Permutations and Combinations)