Let’s analyze and solve each question:

Question 1

How many different 3-digit numbers are possible using only the digits 2, 6, 7, and 9?

Since repetition of digits is allowed:

• We have four options (2, 6, 7, 9) for each digit in the 3-digit number.
• Therefore, the total number of combinations is:

$4 \times 4 \times 4 = 4^3 = 64$

So, 64 different 3-digit numbers can be formed using the digits 2, 6, 7, and 9 with repetition allowed.

Question 2

How many different 3-digit numbers are possible using only the digits 2, 6, 7, and 9, if repetition of digits is not allowed?

Since repetition is not allowed:

• For the first digit, we have 4 choices (2, 6, 7, 9).
• For the second digit, we have only 3 remaining choices (since we can’t repeat the first digit).
• For the third digit, we have 2 remaining choices.

Thus, the total number of combinations is:

$4 \times 3 \times 2 = 24$

So, 24 different 3-digit numbers can be formed using the digits 2, 6, 7, and 9 without repetition.

Would you like further clarification on any part of this?

Here are some related questions to deepen your understanding:

1. How many 3-digit numbers are possible if we use only even digits and repetition is allowed?
2. What if we add a restriction that the first digit cannot be 2, but repetition is allowed?
3. How many 4-digit numbers can be formed with the digits 2, 6, 7, and 9 if repetition is not allowed?
4. How many 3-digit numbers are possible if we add the digit 5 and allow repetition?
5. Can we form a 3-digit number that is a multiple of 3 using only the digits 2, 6, 7, and 9?

Tip: When dealing with permutations and combinations, identifying whether repetition is allowed or not is crucial for determining the correct approach.