How many numbers lying between $100$ and $1000$ can be formed with the digits $0, 1, 2, 3, 4, 5$, if the repetition of the digits is not allowed?

Step 1: Determine the number of digits

The problem asks for numbers lying between $100$ and $1000$. This means we are looking for three-digit numbers. A three-digit number can be represented as $ABC$, where $A$, $B$, and $C$ are digits.

Step 2: Determine choices for the first digit

For a three-digit number $ABC$, the first digit $A$ cannot be $0$, because if $A$ were $0$, it would be a two-digit number. The available digits are $0, 1, 2, 3, 4, 5$. So, $A$ can be any of $1, 2, 3, 4, 5$. This gives $5$ choices for the first digit.

Step 3: Determine choices for the second digit

Since repetition of digits is not allowed, one digit has already been used for $A$. We started with $6$ available digits ($0, 1, 2, 3, 4, 5$). After choosing $A$, $5$ digits remain. Since $B$ can be $0$, there are $5$ choices for the second digit.

Step 4: Determine choices for the third digit

Two digits have already been used for $A$ and $B$. From the initial $6$ digits, $4$ digits remain. Therefore, there are $4$ choices for the third digit $C$.

Step 5: Calculate the total number of numbers

To find the total number of three-digit numbers that can be formed, we multiply the number of choices for each position: $5$ choices for the first digit, $5$ choices for the second digit, and $4$ choices for the third digit.

Step 6: Final Calculation

Multiplying the number of choices for each position gives us the total count: $5 \times 5 \times 4 = 100$.

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