Question 12 of 42beginner🔧 ApplyNumerical2 marks

How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?

Correct Answer

504

Exercise: EXERCISE 6.3 | Q: 1 | (Page 15)
For More Understanding

Explanation

This question follows the same pattern as Example 10 in the textbook, which calculates permutations for forming numbers without repetition. Since we need 3-digit numbers using digits 1 to 9 without repetition, we calculate permutations of 9 items taken 3 at a time. The hundreds place can be filled in 9 ways, tens place in 8 ways, and units place in 7 ways, giving 9×89 \times 8 × 7 = 504.

Solution Steps

  1. Step 1: Identify that we need to form 3-digit numbers using digits 1 to 9 without repetition. Since order matters (e.g., 123 and 132 are different numbers), this is a permutation problem.

  2. Step 2: Apply the permutation formula. Number of 3-digit numbers = 9P₃ = 9!/(9-3)! = 9!/6!

  3. Step 3: Calculate: 9!/6! = 9×89 \times 8 × 7 = 504

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