We are given the digits 5, 3, and 7 and need to determine how many different numbers can be formed using them without repetition.
Step 1: Count Possible Number Lengths
Since we are not allowing repetition, we can form numbers of lengths 1, 2, and 3.
Step 2: Compute the Count for Each Length
1-digit numbers:
We can choose any of the 3 digits: {5, 3, 7}.
So, there are 3 numbers.
2-digit numbers:
We choose 2 digits out of 3 and arrange them.
Number of ways to select and arrange: \[ P(3,2) = \frac{3!}{(3-2)!} = \frac{3!}{1!} = \frac{3 \times 2}{1} = 6. \] 3-digit numbers:
We must use all 3 digits and arrange them.
Number of ways to arrange 3 digits: \[ P(3,3) = 3! = 3 \times 2 \times 1 = 6. \] Step 3: Compute the Total Count
Adding all possibilities together: \[ 3 + 6 + 6 = 15. \] Thus, the total number of numbers that can be formed is: \( 15. \)