ashleynguyenx3
  • ashleynguyenx3
Probability: How many different permutations are there if you used 4 different digits? Repeat a digit?
Mathematics
chestercat
  • chestercat
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unknownunknown
  • unknownunknown
4, 4^2, 4^3, 4^n.... n=repeats
ashleynguyenx3
  • ashleynguyenx3
Wouldn't it be 4 factorial?
ashleynguyenx3
  • ashleynguyenx3
Or is it different?

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unknownunknown
  • unknownunknown
That's if you want unique digits, then yes.
unknownunknown
  • unknownunknown
Ahh I see what you mean. You take one out each time? So 4.. 3.. 2.. 1.. yes that's 4!.
ashleynguyenx3
  • ashleynguyenx3
Would it still be that if order matters?
unknownunknown
  • unknownunknown
If you want certain digits in a specific order, then that would change it.
unknownunknown
  • unknownunknown
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
ashleynguyenx3
  • ashleynguyenx3
I think I'm supposed to find the number or permutations not the probability, so I'm not completely sure.
unknownunknown
  • unknownunknown
Then I'd go with 4!. Order wouldn't matter there.
ashleynguyenx3
  • ashleynguyenx3
Okay, so what about when we repeat a number?
unknownunknown
  • unknownunknown
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
unknownunknown
  • unknownunknown
By repeat, you mean we don't exclude the number we pick, right?
ashleynguyenx3
  • ashleynguyenx3
No this question is more for like iphone security codes. If that makes sense.
ashleynguyenx3
  • ashleynguyenx3
So that's why order matters.
unknownunknown
  • unknownunknown
Okay, can you give me an example of two combinations then, that follow what you require.
ashleynguyenx3
  • ashleynguyenx3
So for 3 digits it would be like 123, 132, 213, 231, 321, 312 if there is no repeating
ashleynguyenx3
  • ashleynguyenx3
With a repeating it would be 112, 121, 211
ashleynguyenx3
  • ashleynguyenx3
I know the one without repeating is 4!
unknownunknown
  • unknownunknown
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?
ashleynguyenx3
  • ashleynguyenx3
Yeah I think I found a site that might help
unknownunknown
  • unknownunknown
Alright, good luck.

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