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CAT 2023 Slot 1 — QA Question 1

Numbers (P&C)Easy

The number of all natural numbers up to 1000 with non-repeating digits is:

Answer & solution

  • A

    648

  • B

    585

  • 738

  • D

    504

Solution

Easy

"Up to 1000" means 11 to 999999 (the four-digit 10001000 has a repeated 00). Split by digit-length — 1-digit, 2-digit, 3-digit — and count each with the multiplication principle, taking care that a leading digit can never be 00.

1

1-digit numbers (11–99): every one trivially has non-repeating digits.

count=9\text{count}=9
2

2-digit numbers   ‾   ‾\underline{\ \ }\,\underline{\ \ }: the tens digit avoids 00; the units digit is any of the 1010 digits except the one already used.

9⏟tens (1–9)×9⏟units (≠tens)=81\underbrace{9}_{\text{tens }(1\text{–}9)}\times\underbrace{9}_{\text{units }(\ne\text{tens})}=81
3

3-digit numbers   ‾   ‾   ‾\underline{\ \ }\,\underline{\ \ }\,\underline{\ \ }: hundreds avoids 00 (99 ways), tens is any digit except hundreds (99 ways, since 00 is now allowed), units is any digit except the two used (88 ways).

9×9×8=6489\times 9\times 8=648
4

Add the three buckets:

9+81+648=7389+81+648=738
738(option c)\mathbf{738}\quad\text{(option c)}

Watch the leading digit: the only difference between the tens place of a 2-digit number (99 ways) and the tens place of a 3-digit number (99 ways too) is which digit is banned — in the 2-digit case it's 00, in the 3-digit case it's the hundreds digit. Same count, different reason.

CAT 2023 Slot 1 QA Q1: The number of all natural numbers up to 1000 with non-repeating digits is: — Solution | TheCATExam