UPSC Prelims 2005 · General Studies
Question 111: Basic Numeracy
A square is divided into 9 identical smaller squares. Six identical balls are to be placed in these smaller squares such that each of the three rows gets at least one ball (one ball in one square only). In how many different ways can this be done?
AnswerThe answer is (d).
Choose 6 of the 9 squares for the identical balls: C(9,6) = 84. Subtract the placements that leave a row empty. If one row is empty, all 6 balls fill the other two rows, which can happen in exactly 1 way for each of the 3 rows, so 3 placements. Two empty rows is impossible. Answer: 84 − 3 = 81.
How you could have got this
No shortcut here: in a “how many” question, ruling out options doesn’t help. Every statement has to be judged.
This follows an unofficial answer: UPSC has not published its key for this year, and if that answer is wrong, so is this path.
Which options go is worked out from UPSC’s key and the options, not by a model.
Why each option is right or wrong
- (a) 27 undercounts the placements.
- (b) 36 undercounts the placements.
- (c) 54 undercounts the placements.
- (d) Correct: 84 − 3 = 81.
Explanation: written by UPSC Grade, not by UPSC.
Source: UPSC, Civil Services (Preliminary) Examination 2005, General Studies. UPSC's own paper isn't online; the words are from a published compilation. The paper on upsc.gov.in.