Five digit numbers are formed using the digits 1, 2, 3, 5, 7 with repetitions and are written in descending order with serial numbers. For example, the number 77777 has serial number 1. Then the serial number of 35337 is
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Answer: 1436
Solution
We need to find how many 5-digit numbers (using digits 1,2,3,5,7 with repetition) are greater than 35337.
Order the digits: 1<2<3<5<7, so in descending order, 7 is rank 1, 5 is rank 2, 3 is rank 3, 2 is rank 4, 1 is rank 5.
Count numbers greater than 35337 by position:
Numbers starting with 5 or 7: 2 × 5⁴ = 1250
Numbers starting with 3, then second digit > 5: Second digit must be 7 (only digit > 5 available). That's 1 × 5³ = 125.
Numbers starting with 35, then third digit > 3: Third digit ∈ {5,7} = 2 choices. That's 2 × 5² = 50.
Numbers starting with 353, then fourth digit > 3: Fourth digit ∈ {5,7} = 2 choices. That's 2 × 5¹ = 10.
Numbers starting with 3533, then fifth digit > 7: No digit > 7 exists = 0.
Total numbers greater than 35337 = 1250 + 125 + 50 + 10 + 0 = 1435.
Serial number of 35337 = 1435 + 1 = 1436.
Order the digits: 1<2<3<5<7, so in descending order, 7 is rank 1, 5 is rank 2, 3 is rank 3, 2 is rank 4, 1 is rank 5.
Count numbers greater than 35337 by position:
Numbers starting with 5 or 7: 2 × 5⁴ = 1250
Numbers starting with 3, then second digit > 5: Second digit must be 7 (only digit > 5 available). That's 1 × 5³ = 125.
Numbers starting with 35, then third digit > 3: Third digit ∈ {5,7} = 2 choices. That's 2 × 5² = 50.
Numbers starting with 353, then fourth digit > 3: Fourth digit ∈ {5,7} = 2 choices. That's 2 × 5¹ = 10.
Numbers starting with 3533, then fifth digit > 7: No digit > 7 exists = 0.
Total numbers greater than 35337 = 1250 + 125 + 50 + 10 + 0 = 1435.
Serial number of 35337 = 1435 + 1 = 1436.
[JEE Main 2023 · 29 Jan · Shift 1]