How many 4 digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9 if no digit is repeated and the number must be divisible by 5?

Introduction / Context:
This is a standard permutations question with two key conditions: the number must be 4 digits long, and it must be divisible by 5. Divisibility by 5 for whole numbers is controlled entirely by the last digit, so the problem becomes a mix of a divisibility rule and a permutation count with no repetition of digits allowed.

Given Data / Assumptions:

• Available digits: 2, 3, 5, 6, 7, 9.
• Digits are distinct.
• We must form 4 digit numbers.
• No digit may be repeated within a number.
• The number must be divisible by 5.

Concept / Approach:
A positive integer is divisible by 5 if and only if its last digit is 0 or 5. Among the allowed digits, only 5 satisfies this condition. Therefore, the units position is fixed as 5. Once the last digit is fixed, the remaining three positions (thousands, hundreds and tens) must be filled with any three distinct digits chosen from the remaining 5 digits. Since order matters for positions within a number, we use permutations, not combinations.

Step-by-Step Solution:
Place 5 in the units position to satisfy divisibility by 5. Remaining digits available for the first three positions: 2, 3, 6, 7, 9 (5 digits). We must select and arrange 3 digits out of these 5 without repetition. Number of ways = 5P3 = 5 * 4 * 3. Compute 5 * 4 = 20. Then 20 * 3 = 60.

Verification / Alternative check:
We can check with position wise reasoning. Thousands place can be any of the 5 remaining digits, hundreds place then has 4 choices, and tens place has 3 choices. Multiplying 5 * 4 * 3 again gives 60 possible numbers. There is no other choice for the units digit, so this count is final. Since 60 is not extremely large and fits with having 6 total digits and a strong restriction at the last position, the result looks numerically sensible.

Why Other Options Are Wrong:

• 48: This usually comes from mistakenly assuming leading digit cannot be 0 (which is irrelevant here) or miscounting one of the positions.
• 36: This is too small and often comes from using 4P3 instead of 5P3 for the first three positions.
• 20: This is just the number 5C3, which counts choices of digits but completely ignores the different possible orders.

Common Pitfalls:
A classic error is forgetting that divisibility by 5 fixes the last digit and then still trying to consider other digits in that place. Another issue is using combinations instead of permutations for number formation. Students also sometimes incorrectly worry about leading zero, which does not apply here because 0 is not among the available digits. Keeping a clear picture of fixed and free positions avoids such confusion.

Final Answer:
The number of 4 digit numbers that can be formed under the given conditions is 60.