UPSC Prelims 2016 · CSAT
Question 72: Basic Numeracy
Four-digit numbers are to be formed using the digits 1, 2, 3 and 4; and none of these four digits are repeated in any manner. Further,
- 1.2 and 3 are not to immediately follow each other
- 2.1 is not to be immediately followed by 3
- 3.4 is not to appear at the last place
- 4.1 is not to appear at the first place
How many different numbers can be formed ?
AnswerUPSC’s answer is (a).
List the 24 arrangements of 1, 2, 3, 4 by first digit and strike out those with 1 first, 4 last, 23 or 32 together, or 13 together. The numbers that survive are 2143, 2431, 3142, 3412, 3421 and 4312, which makes 6. Remember: list systematically by first digit so no case is missed or repeated.
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.
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) Right: exactly six numbers survive the rules.
- (b) Over-counts: only six survive.
- (c) Over-counts: only six survive.
- (d) The count is 6, which is option (a).
Answer: UPSC’s key. Explanation: written by UPSC Grade, not by UPSC.
Source: UPSC, Civil Services (Preliminary) Examination 2016, General Studies Paper II (CSAT), Series A. The paper on upsc.gov.in.