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ashleynguyenx3
 one year ago
Probability:
How many different permutations are there if you used 4 different digits? Repeat a digit?
ashleynguyenx3
 one year ago
Probability: How many different permutations are there if you used 4 different digits? Repeat a digit?

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unknownunknown
 one year ago
Best ResponseYou've already chosen the best response.04, 4^2, 4^3, 4^n.... n=repeats

ashleynguyenx3
 one year ago
Best ResponseYou've already chosen the best response.0Wouldn't it be 4 factorial?

ashleynguyenx3
 one year ago
Best ResponseYou've already chosen the best response.0Or is it different?

unknownunknown
 one year ago
Best ResponseYou've already chosen the best response.0That's if you want unique digits, then yes.

unknownunknown
 one year ago
Best ResponseYou've already chosen the best response.0Ahh I see what you mean. You take one out each time? So 4.. 3.. 2.. 1.. yes that's 4!.

ashleynguyenx3
 one year ago
Best ResponseYou've already chosen the best response.0Would it still be that if order matters?

unknownunknown
 one year ago
Best ResponseYou've already chosen the best response.0If you want certain digits in a specific order, then that would change it.

unknownunknown
 one year ago
Best ResponseYou've already chosen the best response.0Probability 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
 one year ago
Best ResponseYou've already chosen the best response.0I think I'm supposed to find the number or permutations not the probability, so I'm not completely sure.

unknownunknown
 one year ago
Best ResponseYou've already chosen the best response.0Then I'd go with 4!. Order wouldn't matter there.

ashleynguyenx3
 one year ago
Best ResponseYou've already chosen the best response.0Okay, so what about when we repeat a number?

unknownunknown
 one year ago
Best ResponseYou've already chosen the best response.0If 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
 one year ago
Best ResponseYou've already chosen the best response.0By repeat, you mean we don't exclude the number we pick, right?

ashleynguyenx3
 one year ago
Best ResponseYou've already chosen the best response.0No this question is more for like iphone security codes. If that makes sense.

ashleynguyenx3
 one year ago
Best ResponseYou've already chosen the best response.0So that's why order matters.

unknownunknown
 one year ago
Best ResponseYou've already chosen the best response.0Okay, can you give me an example of two combinations then, that follow what you require.

ashleynguyenx3
 one year ago
Best ResponseYou've already chosen the best response.0So for 3 digits it would be like 123, 132, 213, 231, 321, 312 if there is no repeating

ashleynguyenx3
 one year ago
Best ResponseYou've already chosen the best response.0With a repeating it would be 112, 121, 211

ashleynguyenx3
 one year ago
Best ResponseYou've already chosen the best response.0I know the one without repeating is 4!

unknownunknown
 one year ago
Best ResponseYou've already chosen the best response.0So 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
 one year ago
Best ResponseYou've already chosen the best response.0Yeah I think I found a site that might help

unknownunknown
 one year ago
Best ResponseYou've already chosen the best response.0Alright, good luck.
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