UPSC Prelims 2004 · General Studies
Question 110: Basic Numeracy
Nine different letters are to be dropped in three different letter boxes. In how many different ways can this be done?
AnswerThe answer is (b).
Each of the nine different letters can go into any of the three boxes independently, giving 3 x 3 x ... nine times = 3^9 ways. 9^3 would count something else, three choices among nine, and 3^9 - 3 would remove the cases where all letters go into one box, which the question does not forbid.
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 is 3^3, which counts only three letters.
- (b) Correct: 3^9, three choices for each of nine letters.
- (c) 9^3 swaps the base and the exponent.
- (d) 3^9 - 3 wrongly removes the all-in-one-box cases.
Explanation: written by UPSC Grade, not by UPSC.
Source: UPSC, Civil Services (Preliminary) Examination 2004, General Studies. UPSC's own paper isn't online; the words are from a published compilation. The paper on upsc.gov.in.