Mathematical difference between permutation&combination

[Jonathan1218]

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 )!

 

[Dale]

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.

 

[Jonathan1218]

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

 

[FactChecker]

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.

 

[Dale]

However I stated repetitions are allowed,

Oops. My apologies, I missed that.

The standard term is “with replacement”, but I should have noticed and understood as you wrote it

Here is a site which covers both the with and without replacement cases:

https://www.mathsisfun.com/combinatorics/combinations-permutations.html

 

[Stephen Tashi]

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}.

 

[Jonathan1218]

Thank you i get it now :)