# In what format does the "permutations" command in the Combinatorics library compute the permutations?
**URL:** https://discourse.julialang.org/t/in-what-format-does-the-permutations-command-in-the-combinatorics-library-compute-the-permutations/74451
**Category:** General Usage
**Tags:** package, combinatorics
**Created:** [January 12, 2022, 1:24am UTC](https://discourse.julialang.org/t/in-what-format-does-the-permutations-command-in-the-combinatorics-library-compute-the-permutations/74451 "2022-01-12T01:24:06Z")
**Posts on this page:** 3
**Page:** 1
### Author:  [@BinChicken42](https://discourse.julialang.org/u/BinChicken42)
#### Post date: [January 12, 2022, 1:24am UTC](https://discourse.julialang.org/t/in-what-format-does-the-permutations-command-in-the-combinatorics-library-compute-the-permutations/74451/1 "2022-01-12T01:24:06Z")
I’m talking about this command: [[Docstrings · Combinatorics.jl](https://docs.juliahub.com/Combinatorics/AwRuT/1.0.2/autodocs/#Combinatorics.permutations-Tuple%7BAny%7D)], Combinatorics version is 1.0.2, Julia version is 1.7.1
I want to see if the digits in a number (except one digit, called dig) add up to dig. My plan is to create all the permutations of the digits without dig and then add each digit together in each permutation so that is looks something like this (For the number 65231, and dig is 6):
First permutation - 5231, 5+2 is bigger than 6, so it continues to the next permutation
2nd - 2315, 2+3 = 5, 2+3+1 = 6 so it stops the for loop
My question is what format does the permutation command return the permutatations as (I can’t just print it because the command knows permutations get big or something). I’m guessing it’s something like this:
`[[5, 2, 3, 1], [2, 3, 1, 5]...]`
Thankyou for your time
---
### Author:  [@nilshg](https://discourse.julialang.org/u/nilshg)
#### Post date: [January 12, 2022, 6:17am UTC](https://discourse.julialang.org/t/in-what-format-does-the-permutations-command-in-the-combinatorics-library-compute-the-permutations/74451/2 "2022-01-12T06:17:02Z")
`permutations` returns an iterator as documented:
```julia
julia> using Combinatorics
julia> permutations(digits(1235))
Combinatorics.Permutations{Vector{Int64}}([5, 3, 2, 1], 4)
```
That’s just how iterators print - they can’t give you all their contents, as that would mean actually accessing the contents of the iterator, which would run counter to the reason for returning an iterator in the first place (namely that you **don’t** have to access all its elements at once).
You can see what’s in your iterator by `collect`ing it (and you’ll see it’s not that big in this case):
```julia
julia> collect(permutations(digits(1235)))
24-element Vector{Vector{Int64}}:
[5, 3, 2, 1]
[5, 3, 1, 2]
[5, 2, 3, 1]
[5, 2, 1, 3]
[5, 1, 3, 2]
[5, 1, 2, 3]
[3, 5, 2, 1]
[3, 5, 1, 2]
[3, 2, 5, 1]
[3, 2, 1, 5]
[3, 1, 5, 2]
[3, 1, 2, 5]
[2, 5, 3, 1]
[2, 5, 1, 3]
[2, 3, 5, 1]
[2, 3, 1, 5]
[2, 1, 5, 3]
[2, 1, 3, 5]
[1, 5, 3, 2]
[1, 5, 2, 3]
[1, 3, 5, 2]
[1, 3, 2, 5]
[1, 2, 5, 3]
[1, 2, 3, 5]
```
I would also note that generating all permutations here seems a bit wasteful, you might be interested in `combinations` instead:
```julia
julia> collect(combinations(digits(1235)))
15-element Vector{Vector{Int64}}:
[5]
[3]
[2]
[1]
[5, 3]
[5, 2]
[5, 1]
[3, 2]
[3, 1]
[2, 1]
[5, 3, 2]
[5, 3, 1]
[5, 2, 1]
[3, 2, 1]
[5, 3, 2, 1]
```
then your whole problem reduces to
```julia
julia> any(x -> sum(x) == 6, combinations(digits(1235)))
true
```
Note that this still isn’t the most efficient thing you can do, as some sums are still evaluated multiple times here, and the problem in general has more structure that you might be able to exploit for a more efficient solution.
---
### Author:  [@BinChicken42](https://discourse.julialang.org/u/BinChicken42)
#### Post date: [January 12, 2022, 8:27am UTC](https://discourse.julialang.org/t/in-what-format-does-the-permutations-command-in-the-combinatorics-library-compute-the-permutations/74451/3 "2022-01-12T08:27:20Z")
I forgot that recursion existed and that’s a much better idea than using permutations