How many integers between 100 and 999 (3 digits), contain only the digits 2,3,4,5,6,7 and 8, and no digit appears twice?

1. Break Down the Problem

We need to determine how many three-digit integers exist within the range of 100 to 999, using only the digits {2, 3, 4, 5, 6, 7, 8}, ensuring that no digit is repeated within any integer.

2. Relevant Concepts

For a three-digit integer:

• The hundreds place must be filled by any digit from the set {2, 3, 4, 5, 6, 7, 8}.
• The tens and units places must be filled by different digits selected from the remaining available options after the hundreds digit has been chosen.

3. Analysis and Detail

1. Choose the Hundreds Place:

   • We have 7 options for the first digit (hundreds place): {2, 3, 4, 5, 6, 7, 8}.

2. Choose the Tens Place:

   • After selecting a digit for the hundreds place, we have 6 digits left to choose from (since no digit can repeat).

3. Choose the Units Place:

   • After the hundreds and tens places are filled, 5 digits remain for the units place.

Calculation:

The total number of valid three-digit integers can be computed by multiplying the number of choices for each place:

$\text{Total} = (\text{Choices for Hundreds}) \times (\text{Choices for Tens}) \times (\text{Choices for Units})$

Substituting in the choices:

$\text{Total} = 7 \times 6 \times 5 = 210$

4. Verify and Summarize

We have carefully accounted for the choices of digits for each position while ensuring no repetition. The calculations have been verified as follows:

• 7 choices for the hundreds place
• 6 remaining choices for the tens place
• 5 choices left for the units place

Final Answer

The total number of integers between 100 and 999 that contain only the digits {2, 3, 4, 5, 6, 7, 8} with no digit appearing twice is 210.