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4, 4^2, 4^3, 4^n.... n=repeats
Wouldn't it be 4 factorial?
Or is it different?
That's if you want unique digits, then yes.
Ahh I see what you mean. You take one out each time? So 4.. 3.. 2.. 1.. yes that's 4!.
Would it still be that if order matters?
If you want certain digits in a specific order, then that would change it.
Probability of selecting one digit from 4 is 0.25. Then 1 digit from 3 is 0.33, 1 digit from 2 is 0.5, 1 digit from 1 is 1. So the probability of doing all of that correctly is 0.25*0.33*0.5
I think I'm supposed to find the number or permutations not the probability, so I'm not completely sure.
Then I'd go with 4!. Order wouldn't matter there.
Okay, so what about when we repeat a number?
If we repeat, then instead of doing 4 * 3 * 2 *1, we have four digits still in each case. So we do 4 * 4 * 4 * 4 = 4^4 permutation
By repeat, you mean we don't exclude the number we pick, right?
No this question is more for like iphone security codes. If that makes sense.
So that's why order matters.
Okay, can you give me an example of two combinations then, that follow what you require.
So for 3 digits it would be like 123, 132, 213, 231, 321, 312 if there is no repeating
With a repeating it would be 112, 121, 211
I know the one without repeating is 4!
So for any two numbers a and b, a appears twice, and b once, in all three combinations. And do that for all numbers from 1 to 4?
Yeah I think I found a site that might help
Alright, good luck.