To determine how many unique numbers can be formed using the digits 1 through 9, without repeating any digits, let's consider the following:

1. We have 9 distinct digits: 1, 2, 3, 4, 5, 6, 7, 8, and 9.
2. We can create numbers of different lengths (from 1-digit to 9-digit numbers) by arranging these digits in different ways.
3. The number of arrangements (or permutations) for each length can be calculated using factorials.

Solution

• 1-digit numbers: We have 9 options (one for each digit).

  • Total: $9$

• 2-digit numbers: For the first digit, we have 9 choices, and for the second digit, we have 8 choices (since one digit is already used).

  • Total: $9 \times 8 = 72$

• 3-digit numbers: For the first digit, we have 9 choices, the second digit has 8 choices, and the third digit has 7 choices.

  • Total: $9 \times 8 \times 7 = 504$

• 4-digit numbers: For the first four digits, we have choices in the following order: 9, 8, 7, and 6.

  • Total: $9 \times 8 \times 7 \times 6 = 3024$

• 5-digit numbers: For the first five digits, the choices are: 9, 8, 7, 6, and 5.

  • Total: $9 \times 8 \times 7 \times 6 \times 5 = 15120$

• 6-digit numbers: For the first six digits, the choices are: 9, 8, 7, 6, 5, and 4.

  • Total: $9 \times 8 \times 7 \times 6 \times 5 \times 4 = 60480$

• 7-digit numbers: For the first seven digits, the choices are: 9, 8, 7, 6, 5, 4, and 3.

  • Total: $9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 = 181440$

• 8-digit numbers: For the first eight digits, the choices are: 9, 8, 7, 6, 5, 4, 3, and 2.

  • Total: $9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 = 362880$

• 9-digit numbers: For a complete 9-digit arrangement, each digit can be used exactly once.

  • Total: $9! = 362880$

Total Unique Numbers

Adding all these results together:

$9 + 72 + 504 + 3024 + 15120 + 60480 + 181440 + 362880 + 362880 = 951409$

Thus, a total of 951,409 unique numbers can be formed.

Would you like further details on any steps, or have any questions?

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Here are five related questions:

1. How does the concept of permutations apply if some digits could be repeated?
2. What changes if we want only even numbers?
3. How would the solution change if we only wanted prime numbers?
4. What are the possible sums of all these unique numbers?
5. How many numbers are formed if we only allow 3-digit combinations?

Tip: Remember, the factorial function (n!) is a key tool in counting arrangements where order matters and no repeats are allowed.