How to make permutation in bash with N! input?

Question

I have to make a permutation in bash with \"eval\" and \"seq\" commands. So I have to make first a permutation that could contain the same numbers, then I have to filter it

Answer 1

It is a permutation problem, so I have found some others have did it. You can see the answer Generating permutations using bash

So through the answer, you can write the code like this:

perm() {
    local items="$1"
    local out="$2"
    local i
    [[ "$items" == "" ]] && echo "$out" && return
    for (( i=0; i<${#items}; i++ )) ; do
        perm "${items:0:i}${items:i+1}" "$out${items:i:1}"
    done
}

test() {
    local number="$1"
    local iniitem="$(seq -s' ' 1 ${number} | sed -n 's/ //g;p')"
    perm "$iniitem"
}

Then you can use the function like this:
test 3:
the output:

123
132
213
231
312
321