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A company plans to assign identification numbers to its employees. Each number is to consist of four different digits from \(0\) to \(9\), inclusive, except that the first digit cannot be \(0\). How many different identification numbers are possible?
Let's break down what we need to create: identification numbers for employees.
Think of it like creating license plates, but with specific rules:
- Each number must have exactly 4 digits
- All 4 digits must be different (no repeating digits like 1123)
- We can use any digit from 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
- The first digit cannot be 0 (so we won't have numbers like 0123)
Our goal is to count how many different identification numbers we can create following these rules.
Process Skill: TRANSLATE - Converting the word problem into clear mathematical constraints
Let's think about this step by step, imagining we're filling in each position of our 4-digit number: _ _ _ _
For the first position (leftmost digit):
- We cannot use 0
- We can use: 1, 2, 3, 4, 5, 6, 7, 8, 9
- That's 9 possible choices
For the second position:
- We can use any digit from 0-9, BUT
- We cannot repeat the digit we already used in the first position
- Since we used 1 digit already, we have \(10 - 1 = 9\) remaining choices
For the third position:
- We can use any digit from 0-9, BUT
- We cannot repeat either of the 2 digits we've already used
- Since we used 2 digits already, we have \(10 - 2 = 8\) remaining choices
For the fourth position:
- We can use any digit from 0-9, BUT
- We cannot repeat any of the 3 digits we've already used
- Since we used 3 digits already, we have \(10 - 3 = 7\) remaining choices
Process Skill: APPLY CONSTRAINTS - Systematically considering how each restriction affects our choices
Let's work through a concrete example to make this clear:
Suppose we choose 5 for the first position: 5 _ _ _
- First position: 5 (we had 9 choices: 1,2,3,4,5,6,7,8,9)
For the second position, we can use any digit except 5: 5 _ _ _
- Available digits: 0,1,2,3,4,6,7,8,9 (that's 9 choices)
- Let's say we choose 3: 5 3 _ _
For the third position, we can use any digit except 5 and 3: 5 3 _ _
- Available digits: 0,1,2,4,6,7,8,9 (that's 8 choices)
- Let's say we choose 0: 5 3 0 _
For the fourth position, we can use any digit except 5, 3, and 0: 5 3 0 _
- Available digits: 1,2,4,6,7,8,9 (that's 7 choices)
- Let's say we choose 7: 5 3 0 7
This gives us the identification number 5307. The key insight is that no matter which specific digits we choose, the number of available choices for each position follows the same pattern: 9, then 9, then 8, then 7.
Now we can find the total number of possible identification numbers.
The multiplication principle tells us: if we have multiple independent choices to make, we multiply the number of options for each choice.
In our case:
- First position: 9 choices
- Second position: 9 choices
- Third position: 8 choices
- Fourth position: 7 choices
Total number of identification numbers = \(9 \times 9 \times 8 \times 7\)
Let's calculate step by step:
- \(9 \times 9 = 81\)
- \(81 \times 8 = 648\)
- \(648 \times 7 = 4,536\)
Therefore, there are 4,536 different possible identification numbers.
The answer is 4,536 different identification numbers.
Looking at our answer choices, this matches choice B: 4,536.
We can verify this makes sense: we're selecting 4 different digits with the first digit restricted (no 0), so we expect fewer possibilities than if all positions were unrestricted, but still a substantial number given we have 10 digits to work with.
Students often misread this constraint and think that 0 cannot be used anywhere in the 4-digit number. This leads them to believe they can only use digits 1-9 throughout, giving them 9 choices for each position instead of recognizing that 0 can be used in positions 2, 3, and 4.
Many students focus on the "first digit cannot be 0" rule but completely miss that all four digits must be different. They might calculate \(9 \times 10 \times 10 \times 10\), allowing repetition of digits, which significantly overcounts the possibilities.
Some students recognize this as a permutation problem but incorrectly try to apply \(\mathrm{P}(10,4) = \frac{10!}{(10-4)!} = 5,040\), forgetting about the restriction that the first digit cannot be 0. This leads them to the wrong answer choice C.
After correctly identifying 9 choices for the first position, students might think the second position has 10 choices (forgetting that one digit is already used) or think it has only 8 choices (incorrectly assuming 0 is still unavailable). The correct count is 9 choices for the second position.
Even with the correct setup of \(9 \times 9 \times 8 \times 7\), students often make calculation mistakes. Common errors include: \(9 \times 9 = 72\) (instead of 81), or \(81 \times 8 = 648\) is correct but then \(648 \times 7 = 4,446\) (instead of 4,536).
No likely faltering points - the calculation directly leads to 4,536, which clearly matches answer choice B. Students who have executed the approach correctly are unlikely to make selection errors at this stage.