Mathematical difference between permutation&combination

  • #1
I know to calculate a combination with n options and r selections while repetition is allowed you are supposed to do (n+r-1)! / r!(n-r)! and the purpose of r! as a denominator is to act as a ratio to eliminate same outcomes but of different orders and is what makes us calculate the combination instead of a permutation.
However why can I not simply remove the r! to use it to calculate permutation? so: ( n+r-1 )! / ( n-r )!
 

Answers and Replies

  • #2
Dale
Mentor
Insights Author
2020 Award
30,844
7,445
I know to calculate a combination with n options and r selections while repetition is allowed you are supposed to do (n+r-1)! / r!(n-r)!
Your formula here is incorrect. It should be ##\frac{n!}{r! (n-r)!}##

See https://betterexplained.com/articles/easy-permutations-and-combinations/

the purpose of r! as a denominator is to act as a ratio to eliminate same outcomes but of different orders and is what makes us calculate the combination instead of a permutation.
Yes.

However why can I not simply remove the r! to use it to calculate permutation?
You can, but

so: ( n+r-1 )! / ( n-r )!
You have to start with the right formula for combinations, not this.
 
  • Like
Likes pbuk
  • #3
Your formula does not take into accounts that there are possible repetitions, in your link only one person can get first, second and so on. With every medal being awarded the number of candidates decreases. However I stated repetitions are allowed, so an example would be getting 3 scoops of 5 flavours of ice cream. I pick 3 between chocolate, vanilla, banana, strawberry and bubblegum. So I can pick triple chocolate, and the formula has to be slightly modified to account for the extra possibilities of repetitions and having n! will not work because the amount of options does not decrease with each selection. Thank you nevertheless
 
  • #4
FactChecker
Science Advisor
Gold Member
6,052
2,337
The r! in the denominator is how many ways r distinct items can be arranged. Because you are allowing repeats and there are identical items in the r items, getting rid of the r! is the wrong thing to do.
 
  • Like
Likes Jonathan1218 and Dale
  • #6
Stephen Tashi
Science Advisor
7,571
1,465
However why can I not simply remove the r! to use it to calculate permutation? so: ( n+r-1 )! / ( n-r )!
Perhaps your question is:
Since the number of combinations of n things taken r at a time with replacement is ##\frac{ (n+r-1)!} {r! (n-1)!}## then why isn't the number of permutations of n things taken r at a time with replacement equal to ##r! \frac{ (n+r-1)!} {r! (n-1)!} ## ?

For example the ways of picking two things from the set {1,2,3} with replacment allowed are:
{1,1}, {1,2}, {1,3}, {2,2}, {2,3}, {3,3} but not all of those choices can be associated with 2! = 2 permutations. For example {1,1} isn't associated with 2 arrangements. If we could distinguish the two 1's as {1A,1B} then we'd distinguish {1A,1B} and {1B,1A}.
 
  • Like
Likes Jonathan1218 and FactChecker
  • #7
Thank you i get it now :)
 

Related Threads on Mathematical difference between permutation&combination

  • Last Post
2
Replies
28
Views
2K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
6
Views
22K
  • Last Post
Replies
14
Views
8K
Replies
3
Views
6K
Replies
4
Views
913
  • Last Post
Replies
5
Views
806
Replies
2
Views
3K
Replies
2
Views
2K
Replies
1
Views
6K
Top