Possible 3-Digit Numbers using Permutations with Repetition Allowed

To solve this problem, we need to find the number of possible 3-digit numbers that can be formed using permutations with repetition allowed if the digits are 1-9.

Permutations with repetition allowed means that we can use each digit more than once in a number.

We can break down the problem into three steps:

Step 1: Determine the number of choices for each digit

Since we are allowed to use digits 1-9, we have 9 choices for the first digit and each subsequent digit.

Step 2: Determine the number of possible arrangements

Since we are forming a 3-digit number, we need to arrange the digits in a specific order. We can use the formula for permutations to determine the number of possible arrangements:

nPr = n! / (n - r)!

where n is the number of choices and r is the number of items being arranged.

n = 9 (choices)
r = 3 (digits in a number)

9P3 = 9! / (9 - 3)!
9P3 = 9! / 6!
9P3 = (9 x 8 x 7 x 6!) / 6!
9P3 = 9 x 8 x 7
9P3 = 504

Therefore, there are 504 possible arrangements of 3 digits using permutations with repetition allowed.

Step 3: Determine the number of unique 3-digit numbers

Since we are allowed to repeat digits, some of the 504 arrangements may be duplicates. To determine the number of unique 3-digit numbers, we can divide the number of arrangements by the number of arrangements for each digit.

n = 9 (choices)
r = 1 (digit in a number)

9P1 = 9! / (9 - 1)!
9P1 = 9! / 8!
9P1 = 9

There are 9 possible arrangements for each digit.

504 / 9 = 56

Therefore, there are 56 unique 3-digit numbers that can be formed using permutations with repetition allowed if digits are 1-9.

Answer: Option C (729) is incorrect.