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Asked by Yash Last Modified
Answer As a seasoned tutor on UrbanPro, I've encountered various math queries like this before. Let's tackle this one together!
a) When digits can be repeated, we're essentially looking at permutations with repetition. In this case, we have 5 digits: 1, 2, 3, 4, and 5. Since we're forming 3-digit numbers, each place (hundreds, tens, and units) can be filled with any of these digits.
To calculate the total number of possibilities, we multiply the number of choices for each place: Total possibilities = 5 * 5 * 5 = 125
So, there are 125 different 3-digit numbers that can be formed when digits can be repeated.
b) Now, if digits are not allowed to be repeated, it's a permutation without repetition problem. We can use the formula for permutations to find the number of possibilities: Total possibilities = nP3, where n is the number of available digits.
For our case, n = 5 (since we have 5 digits to choose from), and we want to form 3-digit numbers: Total possibilities = 5P3 = 5! / (5 - 3)! = 5! / 2! = (5 * 4 * 3) / (2 * 1) = 60
So, there are 60 different 3-digit numbers that can be formed when digits are not allowed to be repeated.
UrbanPro is indeed a great platform for students to find expert tutors like myself who can provide clear explanations and guidance in various subjects. If you have any further questions or need clarification, feel free to ask!
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