To determine how many numbers can be formed using the digits $1, 2, 3, 4, 5, 6, 7, 8,$ and $9$ with each digit used only once, let's analyze the scenario step-by-step.

Solution

1. Total Digits: We have 9 unique digits (1 to 9).
2. Forming Different Length Numbers: We can form numbers with lengths ranging from 1-digit to 9-digit.
3. Counting Arrangements: For a given length $k$, the number of different $k$-digit numbers that can be formed is given by the number of ways to choose $k$ digits out of 9 and arrange them.

Since each digit can only be used once, the number of possible arrangements for each $k$-digit number can be calculated using permutations.

4. Using Permutations:

   • For a $k$-digit number, the formula for the total arrangements is: $P(9, k) = \frac{9!}{(9 - k)!}$ where $9!$ is the factorial of 9, representing all 9 digits being used, and $(9 - k)!$ adjusts for only using $k$ of the digits.

5. Calculating Each Length:

   • 1-digit: $P(9, 1) = 9$
   • 2-digit: $P(9, 2) = 72$
   • 3-digit: $P(9, 3) = 504$
   • 4-digit: $P(9, 4) = 3024$
   • 5-digit: $P(9, 5) = 15120$
   • 6-digit: $P(9, 6) = 60480$
   • 7-digit: $P(9, 7) = 181440$
   • 8-digit: $P(9, 8) = 362880$
   • 9-digit: $P(9, 9) = 362880$

6. Adding All Possible Numbers: $9 + 72 + 504 + 3024 + 15120 + 60480 + 181440 + 362880 + 362880 = 952409$

Answer

The total number of numbers that can be formed is 952,409.

Would you like a detailed breakdown of any of these steps, or have any questions?

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1. How are permutations different from combinations in these types of problems?
2. What happens if we allow digits to be repeated?
3. How many numbers would be possible if only even-length numbers were allowed?
4. What if we limit the digits to 1-5 instead of 1-9?
5. How can factorial values simplify large counting problems?

Tip: When arranging unique items in all possible orders, permutations without repetition offer the best approach.