From the digits 2, 3, 5, 6, 7 and 9, how many different three digit numbers can be formed that are divisible by 5, if no digit is repeated within a number?

Introduction / Context:
This is a standard permutations question with an additional divisibility condition. We have to form three digit numbers from a given list of digits, with no repetition allowed, and the numbers must be divisible by 5. Divisibility by 5 gives a direct condition on the units digit, which significantly simplifies the counting process.

Given Data / Assumptions:

• Available digits are 2, 3, 5, 6, 7 and 9.
• We must form three digit numbers.
• No digit is repeated within a number.
• The number must be divisible by 5.

Concept / Approach:
A number is divisible by 5 if and only if its units (ones) digit is 0 or 5. Because the digit 0 is not in the given set, the units digit must be 5. Once the units digit is fixed as 5, we choose the hundreds and tens digits from the remaining five digits. Since order of these two positions matters and repetition is not allowed, we use permutations for the remaining positions.

Step-by-Step Solution:
Step 1: Fix the units digit as 5 to satisfy the condition of divisibility by 5.Step 2: The remaining digits available for the hundreds and tens places are {2, 3, 6, 7, 9}, a total of 5 digits.Step 3: For the hundreds place, we can choose any one of these 5 digits (none of them is zero, so leading zeros are not an issue).Step 4: After fixing the hundreds digit, there are 4 remaining digits to choose from for the tens place.Step 5: Therefore, the total number of three digit numbers that can be formed is 5 × 4 = 20.
Verification / Alternative check:
We can manually reason with permutations: the number of ways to arrange any two of the five available digits in order is 5P2 = 5 × 4 = 20, and each such pair, when followed by the fixed digit 5, forms a unique three digit number divisible by 5. This confirms the earlier count of 20.

Why Other Options Are Wrong:
The counts 5, 10, 15 or 25 do not correspond to correct permutations of the digits under the divisibility constraint. For example, 5 would only correspond to fixing the first digit and ignoring the second entirely, which is incorrect. The only count that correctly reflects the choices for both hundreds and tens positions is 20.

Common Pitfalls:
Students sometimes forget the rule for divisibility by 5 and try to consider many units digits. Others miscount by allowing repetition of digits even though the question explicitly restricts it. A clear stepwise approach, first fixing the units digit and then arranging the remaining digits, helps avoid these errors.

Final Answer:
The number of three digit numbers divisible by 5 that can be formed is 20.