Number System Question 8 - CAT 2024

Step-by-Step Explanation

We need to find the count of positive integers up to 500 (1 ≤ N ≤ 500) that contain no repeating digits. Break the calculation down by the number of digits:

• Single-digit numbers (1 to 9):

Any single digit is naturally non-repeating.

Count = 9 numbers

• Two-digit numbers (10 to 99):
  • The tens place can be filled by any digit from 1 to 9 (9 choices, as 0 is invalid).
  • The units place can be filled by any digit from 0 to 9 except the one chosen for the tens place (9 remaining choices).

Count = 9 × 9 = 81 numbers

• Three-digit numbers (100 to 499):
  • The hundreds place can be filled by 1, 2, 3, or 4 (4 choices).
  • The tens place can be filled by any digit from 0 to 9 except the one used in the hundreds place (9 choices).
  • The units place can be filled by any digit from 0 to 9 except the two already used (8 choices).

Count = 4 × 9 × 8 = 288 numbers

• The boundary number (500):

The number 500 contains repeating zeros, so it is invalid.

Total valid positive integers = 9 + 81 + 288 = 378

Correct Answer: 378

💡 FundaMakers Tip:

When counting permutations for numbers, remember that 0 cannot occupy the leading place value. However, 0 becomes a valid option for the second slot, which is why the number of available choices remains 9 for both the first and second slots in two-digit numbers.