Permutation Formula Not Working

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Discussion Overview

The discussion revolves around the calculation of permutations and combinations for three variables (daily, weekly, monthly) that can each take on a value of 'yes' or 'no'. Participants are exploring the correct formula to determine the number of possible 'yes' permutations and clarifying the distinction between permutations and combinations.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant applies the permutation formula n! / (n-r)! and calculates 6, questioning the assumptions behind this approach.
  • Another participant points out that there are actually 8 possible combinations, including the case where all variables are 'no'.
  • A different participant suggests that the correct analysis involves using r^n for total possibilities, stating that in this case, 2^3 equals 8.
  • One participant corrects the initial poster by stating that the problem is not a permutation problem, emphasizing that permutations involve changing the order of items, which is not applicable here.
  • Another participant agrees, noting that the situation described is more aligned with combinations rather than permutations.

Areas of Agreement / Disagreement

Participants express disagreement regarding the appropriate mathematical approach, with some advocating for the use of combinations and others attempting to apply permutation formulas. The discussion remains unresolved as different interpretations of the problem persist.

Contextual Notes

Participants have not reached a consensus on the definitions and applications of permutations versus combinations, leading to confusion in the calculations presented. There are also unresolved assumptions regarding the interpretation of the problem.

John2357
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I am stuck on this. I have three variables (daily, weekly, monthly), they can have a value of yes or no. I need to know the number of possible 'yes' permutations. I use the formula:

n! / (n-r)! which yields '6' i.e., 3*2*1 / (3-3)! --> 6/1 = 6

Yet when I do it by hand:
count Monthly Weekly Daily

1 Yes Yes Yes
2 Yes Yes
3 Yes Yes
4 Yes Yes
5 Yes
6 Yes
7 Yes

By hand I get that there are 7 possible permutations.
My question is the correct formula to use:
(n! / (n-r)! ) + 1 ?

Is there a different formula I should use? What am I assuming incorrectly?
Many thanks!
 
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hmm, the parser messed up my post:

Count...Monthly...Weekly...Daily
1...Yes.....Yes....Yes
2...Yes.....Yes.....
3......Yes....Yes
4...Yes......Yes
5...Yes........
6......Yes.....
7.........Yes
 
First off, there are actually 8 possible options for your problem, not 7. You didn't include No,No,No. Second, you really aren't using the right analysis. When you have n objects, each of which has r possibilities, there are r^n total possibilities. In this case 2^3 = 8. You could use the permutation formula, but you have it wrong, it is actually n! / ((n-r)! r!). In this case you would have to include 3!/0!(3-0)! + 3!/1!(3-1)! + 3!/2!(3-2)! + 3!/3!(3-3)! = 1 + 3 + 3 +1 = 8.
 
Cool, thanks for reminding me - it has been 20 years since I last looked at permutations. Thank you!
 
Your crucial difficulty is that this is NOT a permutation problem! "Permutations" refers to changing the order or position of things- and no change of position is happening here.
 
Yeah Permutations count position like a lock, you're talking about strictly combination
 

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