Understanding the Problem: Forming Numbers Without Repetition

The question asks us to find the total count of unique numbers that can be created using only the digits 5, 3, and 7, with the condition that each digit can be used at most once in any given number (no repetitions allowed).

Identifying Available Digits

We have three distinct digits to work with: 5, 3, and 7.

Determining Possible Number Lengths

Since repetitions are not allowed, we can form numbers using one, two, or all three of these digits.

Calculating Numbers for Each Length

• 1-Digit Numbers:

  We can form numbers using each digit individually. The possible 1-digit numbers are 5, 3, and 7. There are 3 such numbers.

• 2-Digit Numbers:

  To form a 2-digit number, we need to choose and arrange 2 digits out of the 3 available digits (5, 3, 7). This is a permutation problem.

  The number of ways to arrange 2 distinct digits chosen from 3 is calculated using the permutation formula $P(n, k) = \frac{n!}{(n-k)!}$, where $n=3$ (total digits) and $k=2$ (digits to choose).

  Number of 2-digit numbers = $P(3, 2) = \frac{3!}{(3-2)!} = \frac{3!}{1!} = 3 \times 2 \times 1 = 6$.

  Alternatively, we have 3 choices for the first digit and 2 remaining choices for the second digit, giving $3 \times 2 = 6$ possible 2-digit numbers.

• 3-Digit Numbers:

  To form a 3-digit number, we need to arrange all 3 available digits (5, 3, 7). This is also a permutation problem.

  The number of ways to arrange 3 distinct digits out of 3 is calculated as $P(n, k) = \frac{n!}{(n-k)!}$, where $n=3$ and $k=3$.

  Number of 3-digit numbers = $P(3, 3) = \frac{3!}{(3-3)!} = \frac{3!}{0!} = 3 \times 2 \times 1 = 6$. (Note: $0! = 1$)

  Alternatively, we have 3 choices for the first digit, 2 choices for the second, and 1 choice for the third, giving $3 \times 2 \times 1 = 6$ possible 3-digit numbers.

Calculating the Total Number of Numbers

To find the total number of numbers that can be formed, we sum the counts for each possible length:

Total Numbers = (Number of 1-digit numbers) + (Number of 2-digit numbers) + (Number of 3-digit numbers)

Total Numbers = 3 + 6 + 6 = 15

Conclusion

Therefore, there are 15 distinct numbers that can be formed using the digits 5, 3, and 7 without allowing any repetitions.