Question

Probability:
How many different permutations are there if you used 4 different digits? Repeat a digit?

Answer 1

4, 4^2, 4^3, 4^n.... n=repeats

Answer 2

Wouldn't it be 4 factorial?

Answer 3

Or is it different?

Answer 4

That's if you want unique digits, then yes.

Answer 5

Ahh I see what you mean. You take one out each time? So 4.. 3.. 2.. 1.. yes that's 4!.

Answer 6

Would it still be that if order matters?

Answer 7

If you want certain digits in a specific order, then that would change it.

Answer 8

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

Answer 9

I think I'm supposed to find the number or permutations not the probability, so I'm not completely sure.

Answer 10

Then I'd go with 4!. Order wouldn't matter there.

Answer 11

Okay, so what about when we repeat a number?

Answer 12

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

Answer 13

By repeat, you mean we don't exclude the number we pick, right?

Answer 14

No this question is more for like iphone security codes. If that makes sense.

Answer 15

So that's why order matters.

Answer 16

Okay, can you give me an example of two combinations then, that follow what you require.

Answer 17

So for 3 digits it would be like 123, 132, 213, 231, 321, 312 if there is no repeating

Answer 18

With a repeating it would be 112, 121, 211

Answer 19

I know the one without repeating is 4!

Answer 20

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?

Answer 21

Yeah I think I found a site that might help

Answer 22

Alright, good luck.