Using the digits 1, 4, 7, 8 and 9, how many distinct 3 digit numbers can be formed if no digit is repeated in a number?

Introduction / Context:
This question is about forming 3 digit numbers from a set of distinct digits without repetition. Because all the given digits are non zero, any of them may be used in the hundreds place without worrying about leading zeros. The problem is therefore a simple permutation count of selecting and arranging 3 digits out of 5 distinct digits.

Given Data / Assumptions:

• Available digits: 1, 4, 7, 8, 9.
• We want to form 3 digit numbers.
• No digit may repeat within a number.
• All numbers must be valid 3 digit integers (hundreds digit cannot be zero, but there is no zero here anyway).

Concept / Approach:
Because every digit is distinct and all can appear in any position, we simply count permutations of 5 digits taken 3 at a time. This equals P(5, 3) and corresponds to choosing a digit for the hundreds place, then a different digit for the tens place, and then a third different digit for the units place.

Step-by-Step Solution:
Step 1: Consider the hundreds place. There are 5 choices for the first digit (1, 4, 7, 8 or 9). Step 2: Consider the tens place. After choosing the hundreds digit, 4 digits remain available. So there are 4 choices for the tens digit. Step 3: Consider the units place. Now 3 digits remain unused, so there are 3 choices for the units digit. Step 4: Multiply the choices. Total 3 digit numbers = 5 * 4 * 3 = 60.

Verification / Alternative check:
Using permutation notation, the number of such numbers is P(5, 3) = 5! / (5 - 3)! = 5! / 2!. 5! is 120 and 2! is 2, so P(5, 3) = 120 / 2 = 60. This matches the direct step by step counting method, confirming the result.

Why Other Options Are Wrong:
The numbers 26, 50 and 64 do not equal the correct permutation value of 60. They may result from confusion with combinations (which would be C(5, 3) = 10) or from partial counts. The value 40 could come from mistakenly restricting the hundreds digit or failing to consider all remaining choices correctly. None of these values corresponds to 5 * 4 * 3.

Common Pitfalls:
Some learners mistakenly use combinations and compute only the number of ways to choose sets of 3 digits, ignoring order. Others misinterpret the hundreds place restriction when there is no zero among the digits. Remember that for forming numbers where order matters and digits are distinct, permutations are the correct tool. Here, multiplying 5 * 4 * 3 covers all valid arrangements.

Final Answer:
There are 60 distinct 3 digit numbers that can be formed under the given conditions.