Permutation Formula Not Working

  • Thread starter John2357
  • Start date
  • #1
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Main Question or Discussion Point

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!
 

Answers and Replies

  • #2
3
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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
 
  • #3
phyzguy
Science Advisor
4,372
1,352
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.
 
  • #4
3
0
Cool, thanks for reminding me - it has been 20 years since I last looked at permutations. Thank you!
 
  • #5
HallsofIvy
Science Advisor
Homework Helper
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919
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.
 
  • #6
Yeah Permutations count position like a lock, you're talking about strictly combination
 

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