The question is:
Using the numbers $1, 2, 3, 4,$ and $5$, how many $3$-digit combinations can you make? Repetitions are allowed. Thanks!
Edit:
I just watched the video Total 3 Digit Numbers If Repetition allowed & Not allowed - Permutations & Combinations Problems. That answered my question.
But now I'd like to add some limitations, i. e.: I don't want more than $2$ repeats per $3$ combo. So, is it correct to say $5$ (any number) $\times\ 5$ (any number) $\times\ 4$ (any number except the first two) = $100$?
Original question
You can choose any of the 5 numbers as your first digit (5 options). For your second digit, you can also choose any of the 5 because repetitions are allowed. This gives $5\cdot5$ possibilities. Add in a third digit, once again choosing from the 5 numbers, and you have $5\cdot5\cdot5$, or 125, possibilities.
Question with limitations
Consider two cases. In one case, the first two digits are different. There are $5\cdot4$ possibilities that satisfy this. The third can be any of the 5, so there are $5\cdot4\cdot5=100$ possibilities for the first case.
In the second case, the first two digits are the same. There are $5\cdot1=5$ ways this can occur. In this case, the third digit must be different, giving you only 4 options to chose from. This means the second case has $5\cdot1\cdot4=20$ possibilities. Adding together the two cases, there are 120 possibilities.
An alternate way of thinking about the problem with limitations is to work backwards from the problem with no limitations. We have 125 possibilities before introducing limitations. Of those possibilities, there are only 5 that violate the limitations (111, 222, 333, 444, and 555). $125-5=120$, so there are 120 possibilities.