How many distinct 4 digit even numbers can be formed using the digits 2, 7, 5, 3, 9 and 1 if no digit is repeated in a number?

Introduction / Context:
This question tests understanding of how to form even numbers under digit and repetition constraints. We have a fixed set of digits and must form 4 digit numbers that are even and do not repeat digits. Even numbers are characterised by having an even digit in the units place, so we start by deciding which digits can appear at the end and then count valid permutations.

Given Data / Assumptions:

• Available digits: 2, 7, 5, 3, 9, 1.
• We must form 4 digit numbers.
• No digit may repeat within a number.
• A valid number must be even.
• All numbers are in standard decimal notation with no leading zero (and here there is no zero provided).

Concept / Approach:
A number is even if its last digit is even. Among the given digits, only 2 is even. Therefore, for any 4 digit even number using these digits exactly once, the units digit must be 2. Once the units position is fixed, we choose and arrange the remaining three positions (thousands, hundreds and tens) using the remaining five digits, with no repetition. This is a permutation problem for the remaining positions.

Step-by-Step Solution:
Step 1: Identify even digits in the given set. Digits: 2, 7, 5, 3, 9, 1. Only 2 is even. Step 2: Fix the units position. To make the number even, the units digit must be 2. Step 3: Determine the remaining digits and positions. Remaining digits to fill the first three positions: 7, 5, 3, 9, 1 (5 digits). Positions to be filled: thousands, hundreds, tens (3 positions). Step 4: Count the permutations of these 5 digits taken 3 at a time. Number of ways = P(5, 3) = 5 * 4 * 3 = 60. Each such arrangement, followed by the fixed trailing digit 2, gives a unique 4 digit even number.

Verification / Alternative check:
We can confirm the logic by noting that there are 5 choices for the first digit, 4 choices for the second digit and 3 choices for the third digit, since we cannot reuse digits. After choosing these three, the last digit 2 is fixed. Multiplying 5 * 4 * 3 equals 60, which matches the permutation calculation and confirms the result.

Why Other Options Are Wrong:
Values such as 59, 61 and 64 do not match the clean product 5 * 4 * 3. They may arise from small arithmetic errors or an incorrect assumption that there are more or fewer valid last digits. The value 48 would result from incorrectly treating the problem as permutations of only 4 digits from 5 without accounting properly for the fixed units digit or omitting some choices.

Common Pitfalls:
A common mistake is to assume that more than one digit in the set is even or to forget that only the last digit determines evenness. Another frequent error is to treat the first digit differently for fear of a leading zero, but zero is not present in the digits, so there is no restriction beyond non repetition. Correctly fixing the last digit as 2 and then permuting the remaining digits avoids these pitfalls.

Final Answer:
There are 60 distinct 4 digit even numbers that can be formed under the given conditions.