UPSC Prelims 2021·CSAT·Quantitative Aptitude·Combinatorics and Probability

Using 2, 2, 3, 3, 3 as digits, how many distinct numbers greater than 30000 can be formed?

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Last updated 23 May 2026, 3:31 pm IST
  1. A3
  2. B6Correct
  3. C9
  4. D12

Explanation

To form a number greater than 30000 using the digits 2, 2, 3, 3, 3, the number must be a 5-digit number, and its first digit must be 3. 1. The first digit is fixed as 3. (d1 = 3) 2. We are left with the remaining 4 digits to arrange: 2, 2, 3, 3. 3. We need to find the number of distinct permutations of these 4 digits. The formula for permutations with repetitions is n! / (n1! * n2! * ...), where n is the total number of items, and n1, n2, etc., are the counts of repeated items. 4. Here, n = 4 (for the remaining 4 positions). The digit 2 appears 2 times (n1 = 2). The digit 3 appears 2 times (n2 = 2). 5. Number of distinct arrangements = 4! / (2! * 2!) = (4 * 3 * 2 * 1) / ((2 * 1) * (2 * 1)) = 24 / (2 * 2) = 24 / 4 = 6 Therefore, 6 distinct numbers greater than 30000 can be formed. The correct answer is B) 6.
Quantitative Aptitude: Using 2, 2, 3, 3, 3 as digits, how many distinct numbers greater than 30000 can be formed?
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