Permutation Formula Not Working

Click For Summary
The discussion revolves around calculating the number of possible permutations for three variables (daily, weekly, monthly) that can each be either "yes" or "no." The initial formula used, n! / (n-r)!, incorrectly suggests there are 6 permutations. However, the correct approach reveals there are actually 8 combinations when considering all possibilities, including "no" for each variable. The confusion stems from misclassifying the problem as a permutation issue rather than a combination one, as the order of the variables does not change. The accurate calculation involves using the formula for combinations, leading to the conclusion that there are 2^3 = 8 total outcomes.
John2357
Messages
3
Reaction score
0
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!
 
Mathematics news on Phys.org
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
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
Replies
8
Views
2K