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UPSC CSE CSAT 2016 · Question 72 · Counting, permutations & probability

Four-digit numbers are to be formed using the digits 1, 2, 3 and 4; and none of these four digits…

CSAT 2016 · Q72

Counting, permutations & probability Medium

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. 1.2 and 3 are not to immediately follow each other
  2. 2.1 is not to be immediately followed by 3
  3. 3.4 is not to appear at the last place
  4. 4.1 is not to appear at the first place

How many different numbers can be formed?

Answer & explanation

Answer: (a) 6

Listing the arrangements by first digit (1 cannot lead) and striking out any with 2 and 3 side by side, with 1 directly before 3, or with 4 at the end leaves 2143, 2431, 3142, 3412, 3421 and 4312 — six numbers.

  1. 1 cannot come first, so the number starts with 2, 3 or 4.
  2. Starting with 2: 2143 and 2431 work (2134 and 2413 contain 13; 2314 and 2341 contain 23).
  3. Starting with 3: 3142, 3412 and 3421 work (3124 ends in 4; 3214 and 3241 contain 32).
  4. Starting with 4: only 4312 works (4123 and 4231 contain 23; 4132 and 4213 contain 13; 4321 contains 32).
  5. Total = 2 + 3 + 1 = 6.

Remember · With only 24 arrangements, list them systematically by first digit and strike out rule-breakers rather than trusting a formula with overlapping conditions.

Question and answer: UPSC's official GS Paper II (2016, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). ·

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