Question:

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

Updated On: Aug 8, 2024
  • 648
  • 585
  • 504
  • 738
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The Correct Option is D

Approach Solution - 1

The correct option is (D): 738

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Approach Solution -2

 

1. For 1-digit numbers: There are 9 one-digit numbers (from 1 to 9).

2. For 2-digit numbers:
  - For the tens place, any digit from 1 to 9 can be chosen (excluding 0).
  - For the units place, any digit can be chosen.
  - So, there are \(9 \times 9 = 81\) possibilities.

3. For 3-digit numbers:
  - For the hundreds place, any digit from 1 to 9 can be chosen.
  - For the tens place, any digit can be chosen (excluding the digit already used in the hundreds place).
  - For the units place, any remaining digit can be chosen.
  - So, there are \(9 \times 9 \times 8 = 648\) possibilities.

Therefore, the total number of natural numbers up to 1000 with non-repeating digits is:

\[9 + 81 + 648 = 738\]

Thank you for the clarification. The correct answer is indeed \(\boxed{738}\).

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