UPSC Prelims 2007 Q135: Science & Technology – Quantitative Aptitude
In the figure shown above, what is the maximum number of different ways in which 8
identical balls can be placed in small triangles 1, 2, 3 and 4 such that each triangle contains at
least one ball?
- (a) 32
- (b) 35
- (c) 44
- (d) 56
Answer: (b) 35
Explanation
8 identical balls in 4 triangles, each with at least one
This is a "stars and bars" problem.
Step 1: Put 1 ball in each triangle first → 4 balls used, 4 balls left.
Step 2: Share 4 identical balls among 4 triangles freely (zero allowed).
Formula: ways to share n identical items among r groups, each ≥ 1 = C(n − 1, r − 1).
| Method | Formula | Value |
|---|---|---|
| Each triangle ≥ 1 (8 balls, 4 triangles) | C(8 − 1, 4 − 1) = C(7, 3) | 35 |
| After pre-filling (4 balls, zero allowed) | C(4 + 4 − 1, 4 − 1) = C(7, 3) | 35 |
Answer: (b) 35.
Why each option is right or wrong
- (a) Incorrect. 32 does not come from any correct formula here.
- (b) CORRECT. C(7, 3) = 35 ways.
- (c) Incorrect. 44 is not a valid stars-and-bars count for these numbers.
- (d) Incorrect. 56 = C(8, 3) — uses n instead of n − 1.
Key point
Identical items into r groups, each ≥ 1: C(n−1, r−1). Here C(7,3) = 35.
Remember
8 identical balls, 4 boxes, none empty = C(7,3) = 35.
Common mistakes
- Using C(n, r−1) = C(8,3) = 56 — off by one.
- Treating balls as different — they are identical, so order does not matter.
- Forgetting the 'at least one' condition — without it the answer would be C(11,3) = 165.
Exam tip
At least one each → C(n−1, r−1). Zero allowed → C(n+r−1, r−1).
Sources
- NCERT Class XI — Mathematics, Ch 6 (Permutations and Combinations)
- R. S. Aggarwal — Quantitative Aptitude for Competitive Examinations (Permutations and Combinations)