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?
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Step-by-Step Solution
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: 5×5×4=100.