UPSC Prelims 2021 · CSAT
Question 66: Basic Numeracy
Using 2, 2, 3, 3, 3 as digits, how many distinct numbers greater than 30000 can be formed ?
AnswerUPSC’s answer is (b).
A number greater than 30000 made from the digits 2, 2, 3, 3, 3 must start with 3, since a leading 2 gives a number under 30000. With the first digit 3, the remaining digits 2, 2, 3, 3 can be arranged in 4! ÷ (2! × 2!) = 6 ways. All of these exceed 30000, so there are 6 numbers. Remember: for repeated digits, divide the factorial by the factorial of each repeat.
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) 3 is too few; the four remaining digits give 6 arrangements.
- (b) Right: first digit 3, then 4! ÷ (2! × 2!) = 6 arrangements of 2, 2, 3, 3.
- (c) 9 overcounts; the 4 numbers starting with 2 are below 30000.
- (d) 12 overcounts; there are only 10 distinct arrangements in all, and 4 start with 2.
Answer: UPSC’s key. Explanation: written by UPSC Grade, not by UPSC.
Source: UPSC, Civil Services (Preliminary) Examination 2021, General Studies Paper II (CSAT), Series A. The paper on upsc.gov.in.