Permutations Of Digits
The number of distinct four-digit numbers greater than 3000 that can be formed using the digits 1, 2, 3, 4, 5 without repetition of any digit equals which value?
Select the correct option:
Solution
72
Forming numbers with a size restriction is handled by constraining the leading digit first, then arranging the rest, a frequent JEE Advanced pattern. A four-digit number using the distinct digits 1, 2, 3, 4, 5 without repetition must exceed 3000, so its thousands digit must be at least 3, giving 3 choices: 3, 4, or 5. After fixing the leading digit, the remaining three places are filled by any 3 of the remaining 4 digits in P(4,3) = 4·3·2 = 24 ways. By the multiplication principle, the total is 3 × 24 = 72. Option 120 = 5·4·3·2 ignores the size restriction. Option 60 uses an incorrect leading-digit count. Option 96 overcounts the leading choices. Hence there are 72 such numbers. Plausibility check: the unrestricted count of four-digit numbers from these digits is P(5,4) = 120, and exactly three of the five possible leading digits qualify, so the restricted count is (3/5)·120 = 72, consistent with the leading-digit argument. Magnitude conditions on formed numbers are managed by first fixing the most significant digit to satisfy the size constraint, then freely arranging the remaining digits. This leading-digit-first strategy generalizes to divisibility and range problems, where the controlling position is identified and constrained before the multiplication principle is applied to the comparatively free remaining places.
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About This Question
- Subject
- mathematics
- Chapter
- permutations and combinations
- Topic
- permutations of digits
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
72
Forming numbers with a size restriction is handled by constraining the leading digit first, then arranging the rest, a frequent JEE Advanced pattern. A four-digit number using the distinct digits 1, 2, 3, 4, 5 without repetition must exceed 3000, so its thousands digit must be at least 3, giving 3 choices: 3, 4, or 5. After fixing the leading digit, the remaining three places are filled by any 3 of the remaining 4 digits in P(4,3) = 4·3·2 = 24 ways. By the multiplication principle, the total is 3 × 24 = 72. Option 120 = 5·4·3·2 ignores the size restriction. Option 60 uses an incorrect leading-digit count. Option 96 overcounts the leading choices. Hence there are 72 such numbers. Plausibility check: the unrestricted count of four-digit numbers from these digits is P(5,4) = 120, and exactly three of the five possible leading digits qualify, so the restricted count is (3/5)·120 = 72, consistent with the leading-digit argument. Magnitude conditions on formed numbers are managed by first fixing the most significant digit to satisfy the size constraint, then freely arranging the remaining digits. This leading-digit-first strategy generalizes to divisibility and range problems, where the controlling position is identified and constrained before the multiplication principle is applied to the comparatively free remaining places.
This medium difficulty mathematics question is from the chapter permutations and combinations, covering the topic of permutations of digits. It appeared in the 2025 exam.
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