Permutations with some identical elements: Difference between revisions

From Rosetta Code
Content added Content deleted
(→‎{{header|Wren}}: Now uses Wren-perm module.)
Line 404: Line 404:
["AABBBC", "AABBCB", "AABCBB", "AACBBB", "ABABBC", "ABABCB", "ABACBB", "ABBABC", "ABBACB", "ABBBAC", "ABBBCA", "ABBCAB", "ABBCBA", "ABCABB", "ABCBAB", "ABCBBA", "ACABBB", "ACBABB", "ACBBAB", "ACBBBA", "BAABBC", "BAABCB", "BAACBB", "BABABC", "BABACB", "BABBAC", "BABBCA", "BABCAB", "BABCBA", "BACABB", "BACBAB", "BACBBA", "BBAABC", "BBAACB", "BBABAC", "BBABCA", "BBACAB", "BBACBA", "BBBAAC", "BBBACA", "BBBCAA", "BBCAAB", "BBCABA", "BBCBAA", "BCAABB", "BCABAB", "BCABBA", "BCBAAB", "BCBABA", "BCBBAA", "CAABBB", "CABABB", "CABBAB", "CABBBA", "CBAABB", "CBABAB", "CBABBA", "CBBAAB", "CBBABA", "CBBBAA"]
["AABBBC", "AABBCB", "AABCBB", "AACBBB", "ABABBC", "ABABCB", "ABACBB", "ABBABC", "ABBACB", "ABBBAC", "ABBBCA", "ABBCAB", "ABBCBA", "ABCABB", "ABCBAB", "ABCBBA", "ACABBB", "ACBABB", "ACBBAB", "ACBBBA", "BAABBC", "BAABCB", "BAACBB", "BABABC", "BABACB", "BABBAC", "BABBCA", "BABCAB", "BABCBA", "BACABB", "BACBAB", "BACBBA", "BBAABC", "BBAACB", "BBABAC", "BBABCA", "BBACAB", "BBACBA", "BBBAAC", "BBBACA", "BBBCAA", "BBCAAB", "BBCABA", "BBCBAA", "BCAABB", "BCABAB", "BCABBA", "BCBAAB", "BCBABA", "BCBBAA", "CAABBB", "CABABB", "CABBAB", "CABBBA", "CBAABB", "CBABAB", "CBABBA", "CBBAAB", "CBBABA", "CBBBAA"]
</pre>
</pre>

=={{header|Mathematica}}/{{header|Wolfram Language}}==
<lang Mathematica>ClearAll[PermsFromFrequencies]
PermsFromFrequencies[l_List] := Module[{digs},
digs = Flatten[MapIndexed[ReverseApplied[ConstantArray], l]];
Permutations[digs]
]
PermsFromFrequencies[{2, 3, 1}] // Column</lang>
{{out}}
<pre>{1,1,2,2,2,3}
{1,1,2,2,3,2}
{1,1,2,3,2,2}
{1,1,3,2,2,2}
{1,2,1,2,2,3}
{1,2,1,2,3,2}
{1,2,1,3,2,2}
{1,2,2,1,2,3}
{1,2,2,1,3,2}
{1,2,2,2,1,3}
{1,2,2,2,3,1}
{1,2,2,3,1,2}
{1,2,2,3,2,1}
{1,2,3,1,2,2}
{1,2,3,2,1,2}
{1,2,3,2,2,1}
{1,3,1,2,2,2}
{1,3,2,1,2,2}
{1,3,2,2,1,2}
{1,3,2,2,2,1}
{2,1,1,2,2,3}
{2,1,1,2,3,2}
{2,1,1,3,2,2}
{2,1,2,1,2,3}
{2,1,2,1,3,2}
{2,1,2,2,1,3}
{2,1,2,2,3,1}
{2,1,2,3,1,2}
{2,1,2,3,2,1}
{2,1,3,1,2,2}
{2,1,3,2,1,2}
{2,1,3,2,2,1}
{2,2,1,1,2,3}
{2,2,1,1,3,2}
{2,2,1,2,1,3}
{2,2,1,2,3,1}
{2,2,1,3,1,2}
{2,2,1,3,2,1}
{2,2,2,1,1,3}
{2,2,2,1,3,1}
{2,2,2,3,1,1}
{2,2,3,1,1,2}
{2,2,3,1,2,1}
{2,2,3,2,1,1}
{2,3,1,1,2,2}
{2,3,1,2,1,2}
{2,3,1,2,2,1}
{2,3,2,1,1,2}
{2,3,2,1,2,1}
{2,3,2,2,1,1}
{3,1,1,2,2,2}
{3,1,2,1,2,2}
{3,1,2,2,1,2}
{3,1,2,2,2,1}
{3,2,1,1,2,2}
{3,2,1,2,1,2}
{3,2,1,2,2,1}
{3,2,2,1,1,2}
{3,2,2,1,2,1}
{3,2,2,2,1,1}</pre>


=={{header|MiniZinc}}==
=={{header|MiniZinc}}==

Revision as of 21:49, 2 August 2022

Permutations with some identical elements is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

Sometimes you want to find all permutations of elements where some elements are repeated, e.g. you have 3 red balls, 2 blue balls and one black ball.

If you just do all permutations of the 6 elements, each permutation will be duplicated 12 times where you can't tell that the identical elements have switched places.


Given an input of the form   [a1, a2, ···, ak]   where   ak   denotes how many duplicates of element   k   you should have,
each   ak > 0   and the sum of all   ak   is   n.

You should get   n! / (a1! × a2! ... × ak!)   permutations as a result.

(You may, of course, denote the elements   0..k-1   if that works better.)


For example, the input   [2,1]   should give results   (1,1,2),   (1,2,1)   and  (2,1,1).

Alternatively, if zero-based:   (0,0,1),   (0,1,0)   and   (1,0,0).


Task

List the permutations you get from the input   [2, 3, 1].

Optionally output the permutations as strings where the first element is represented by A, the second by B and the third by C  
(the example result would then be AAB, ABA and BAA).


Related tasks



11l

Translation of: Nim

<lang 11l>F shouldSwap(s, start, curr) 

  L(i) start .< curr
     I s[i] == s[curr]
        R 0B
  R 1B

F findPerms(ss, index, n, &res) 

  I index >= n
     res.append(ss)
     R

  V s = copy(ss)
  L(i) index .< n
     I shouldSwap(s, index, i)
        swap(&s[index], &s[i])
        findPerms(s, index + 1, n, &res)
        swap(&s[index], &s[i])

F createSlice(nums, charSet) 

  V result = ‘’
  L(n) nums
     V i = L.index
     L 0 .< n
        result ‘’= charSet[i]
  R result

[String] res1, res2, res3 V nums = [2, 1]

V s = createSlice(nums, ‘12’) findPerms(s, 0, s.len, &res1) print(res1.join(‘ ’)) print()

nums = [2, 3, 1] s = createSlice(nums, ‘123’) findPerms(s, 0, s.len, &res2) L(val) res2 

  print(val, end' I (L.index + 1) % 10 == 0 {"\n"} E ‘ ’)

print()

s = createSlice(nums, ‘ABC’) findPerms(s, 0, s.len, &res3) L(val) res3 

  print(val, end' I (L.index + 1) % 10 == 0 {"\n"} E ‘ ’)</lang>
Output:
112 121 211

112223 112232 112322 113222 121223 121232 121322 122123 122132 122213
122231 122321 122312 123221 123212 123122 132221 132212 132122 131222
211223 211232 211322 212123 212132 212213 212231 212321 212312 213221
213212 213122 221123 221132 221213 221231 221321 221312 222113 222131
222311 223121 223112 223211 231221 231212 231122 232121 232112 232211
312221 312212 312122 311222 321221 321212 321122 322121 322112 322211

AABBBC AABBCB AABCBB AACBBB ABABBC ABABCB ABACBB ABBABC ABBACB ABBBAC
ABBBCA ABBCBA ABBCAB ABCBBA ABCBAB ABCABB ACBBBA ACBBAB ACBABB ACABBB
BAABBC BAABCB BAACBB BABABC BABACB BABBAC BABBCA BABCBA BABCAB BACBBA
BACBAB BACABB BBAABC BBAACB BBABAC BBABCA BBACBA BBACAB BBBAAC BBBACA
BBBCAA BBCABA BBCAAB BBCBAA BCABBA BCABAB BCAABB BCBABA BCBAAB BCBBAA
CABBBA CABBAB CABABB CAABBB CBABBA CBABAB CBAABB CBBABA CBBAAB CBBBAA

Dart

Translation of: Tailspin

<lang dart> import 'dart:io';

void main() { 

 stdout.writeln(distinctPerms([2,3,1]).map((p) => alpha("ABC", p)).toList());

}

String alpha(String alphabet, List<int> perm) { 

 return perm.map((i) => alphabet[i]).join("");

}

Iterable<List<int>> distinctPerms(List<int> reps) sync* { 

 Iterable<List<int>> perms(List<List<int>> elements) sync* {
   if (elements.length == 1) {
     yield List.of(elements[0]);
   } else {
     for (int k = 0; k < elements.length; k++) {
       List<List<int>> tail = [];
       for (int i = 0; i < elements.length; i++) {
         if (i == k) {
           if (elements[i].length > 1) {
             tail.add(List.of(elements[i].skip(1)));
           }
         } else {
           tail.add(elements[i]);
         }
       }
       yield* perms(tail).map((t) {
         t.insert(0, elements[k][0]);
         return t;
       });
     }
   }
 }
 List<List<int>> elements = [];
 for (int i = 0; i < reps.length; i++) {
   elements.add(List.filled(reps[i], i));
 }
 yield* perms(elements);

} </lang>

Output:
[AABBBC, AABBCB, AABCBB, AACBBB, ABABBC, ABABCB, ABACBB, ABBABC, ABBACB, ABBBAC, ABBBCA, ABBCAB, ABBCBA, ABCABB, ABCBAB, ABCBBA, ACABBB, ACBABB, ACBBAB, ACBBBA, BAABBC, BAABCB, BAACBB, BABABC, BABACB, BABBAC, BABBCA, BABCAB, BABCBA, BACABB, BACBAB, BACBBA, BBAABC, BBAACB, BBABAC, BBABCA, BBACAB, BBACBA, BBBAAC, BBBACA, BBBCAA, BBCAAB, BBCABA, BBCBAA, BCAABB, BCABAB, BCABBA, BCBAAB, BCBABA, BCBBAA, CAABBB, CABABB, CABBAB, CABBBA, CBAABB, CBABAB, CBABBA, CBBAAB, CBBABA, CBBBAA]

Factor

Removing duplicates from the list of all permutations: <lang factor>USING: arrays grouping math math.combinatorics prettyprint sequences sets ;

distinct-permutations ( seq -- seq )
   [ CHAR: A + <array> ] map-index "" concat-as <permutations>
   members ;

{ 2 3 1 } distinct-permutations 10 group simple-table.</lang>

Output:
AABBBC AABBCB AABCBB AACBBB ABABBC ABABCB ABACBB ABBABC ABBACB ABBBAC
ABBBCA ABBCAB ABBCBA ABCABB ABCBAB ABCBBA ACABBB ACBABB ACBBAB ACBBBA
BAABBC BAABCB BAACBB BABABC BABACB BABBAC BABBCA BABCAB BABCBA BACABB
BACBAB BACBBA BBAABC BBAACB BBABAC BBABCA BBACAB BBACBA BBBAAC BBBACA
BBBCAA BBCAAB BBCABA BBCBAA BCAABB BCABAB BCABBA BCBAAB BCBABA BCBBAA
CAABBB CABABB CABBAB CABBBA CBAABB CBABAB CBABBA CBBAAB CBBABA CBBBAA

Generating distinct permutations directly (more efficient in time and space):

Translation of: Go

<lang factor>USING: arrays io kernel locals math math.ranges sequences ; IN: rosetta-code.distinct-permutations

should-swap? ( start curr seq -- ? )
   [ nipd nth ] [ <slice> member? not ] 3bi ;
.find-permutations ( seq index n -- )
   index n >= [ seq write bl ] [
       index n [a,b) [
           :> i
           index i seq should-swap? [
               index i seq exchange
               seq index 1 + n .find-permutations
               index i seq exchange
           ] when
       ] each
   ] if ;
first-permutation ( nums charset -- seq )
   [ <array> ] 2map "" concat-as ;
.distinct-permutations ( nums charset -- )
   first-permutation 0 over length .find-permutations nl ;
main ( -- )
   { 2 1 } "12"
   { 2 3 1 } "123"
   { 2 3 1 } "ABC"
   [ .distinct-permutations ] 2tri@ ;

MAIN: main</lang>

Output:
112 121 211 
112223 112232 112322 113222 121223 121232 121322 122123 122132 122213 122231 122321 122312 123221 123212 123122 132221 132212 132122 131222 211223 211232 211322 212123 212132 212213 212231 212321 212312 213221 213212 213122 221123 221132 221213 221231 221321 221312 222113 222131 222311 223121 223112 223211 231221 231212 231122 232121 232112 232211 312221 312212 312122 311222 321221 321212 321122 322121 322112 322211 
AABBBC AABBCB AABCBB AACBBB ABABBC ABABCB ABACBB ABBABC ABBACB ABBBAC ABBBCA ABBCBA ABBCAB ABCBBA ABCBAB ABCABB ACBBBA ACBBAB ACBABB ACABBB BAABBC BAABCB BAACBB BABABC BABACB BABBAC BABBCA BABCBA BABCAB BACBBA BACBAB BACABB BBAABC BBAACB BBABAC BBABCA BBACBA BBACAB BBBAAC BBBACA BBBCAA BBCABA BBCAAB BBCBAA BCABBA BCABAB BCAABB BCBABA BCBAAB BCBBAA CABBBA CABBAB CABABB CAABBB CBABBA CBABAB CBAABB CBBABA CBBAAB CBBBAA

Go

This is based on the C++ code here. <lang go>package main

import "fmt"

func shouldSwap(s []byte, start, curr int) bool { 

   for i := start; i < curr; i++ {
       if s[i] == s[curr] {
           return false
       }
   }
   return true

}

func findPerms(s []byte, index, n int, res *[]string) { 

   if index >= n {
       *res = append(*res, string(s))
       return
   }
   for i := index; i < n; i++ {
       check := shouldSwap(s, index, i)
       if check {
           s[index], s[i] = s[i], s[index]
           findPerms(s, index+1, n, res)
           s[index], s[i] = s[i], s[index]
       }
   }

}

func createSlice(nums []int, charSet string) []byte { 

   var chars []byte
   for i := 0; i < len(nums); i++ {
       for j := 0; j < nums[i]; j++ {
           chars = append(chars, charSet[i])
       }
   }
   return chars

}

func main() { 

   var res, res2, res3 []string
   nums := []int{2, 1}
   s := createSlice(nums, "12")
   findPerms(s, 0, len(s), &res)
   fmt.Println(res)
   fmt.Println()

   nums = []int{2, 3, 1}
   s = createSlice(nums, "123")
   findPerms(s, 0, len(s), &res2)
   fmt.Println(res2)
   fmt.Println()

   s = createSlice(nums, "ABC")
   findPerms(s, 0, len(s), &res3)
   fmt.Println(res3)

}</lang>

Output:
[112 121 211]

[112223 112232 112322 113222 121223 121232 121322 122123 122132 122213 122231 122321 122312 123221 123212 123122 132221 132212 132122 131222 211223 211232 211322 212123 212132 212213 212231 212321 212312 213221 213212 213122 221123 221132 221213 221231 221321 221312 222113 222131 222311 223121 223112 223211 231221 231212 231122 232121 232112 232211 312221 312212 312122 311222 321221 321212 321122 322121 322112 322211]

[AABBBC AABBCB AABCBB AACBBB ABABBC ABABCB ABACBB ABBABC ABBACB ABBBAC ABBBCA ABBCBA ABBCAB ABCBBA ABCBAB ABCABB ACBBBA ACBBAB ACBABB ACABBB BAABBC BAABCB BAACBB BABABC BABACB BABBAC BABBCA BABCBA BABCAB BACBBA BACBAB BACABB BBAABC BBAACB BBABAC BBABCA BBACBA BBACAB BBBAAC BBBACA BBBCAA BBCABA BBCAAB BBCBAA BCABBA BCABAB BCAABB BCBABA BCBAAB BCBBAA CABBBA CABBAB CABABB CAABBB CBABBA CBABAB CBAABB CBBABA CBBAAB CBBBAA]

Haskell

<lang haskell>permutationsSomeIdentical :: [(a, Int)] -> a permutationsSomeIdentical [] = [[]] permutationsSomeIdentical xs = 

 [ x : ys
 | (x, xs_) <- select xs 
 , ys <- permutationsSomeIdentical xs_ ]
 where
   select [] = []
   select ((x, n):xs) =
     (x, xs_) :
     [ (y, (x, n) : cs)
     | (y, cs) <- select xs ]
     where
       xs_
         | 1 == n = xs
         | otherwise = (x, n - 1) : xs

main :: IO () main = do 

 print $ permutationsSomeIdentical [(1, 2), (2, 1)]
 print $ permutationsSomeIdentical [(1, 2), (2, 3), (3, 1)]
 print $ permutationsSomeIdentical [('A', 2), ('B', 3), ('C', 1)]</lang>
Output:
[[1,1,2],[1,2,1],[2,1,1]]
[[1,1,2,2,2,3],[1,1,2,2,3,2],[1,1,2,3,2,2],[1,1,3,2,2,2],[1,2,1,2,2,3],[1,2,1,2,3,2],[1,2,1,3,2,2],[1,2,2,1,2,3],[1,2,2,1,3,2],[1,2,2,2,1,3],[1,2,2,2,3,1],[1,2,2,3,1,2],[1,2,2,3,2,1],[1,2,3,1,2,2],[1,2,3,2,1,2],[1,2,3,2,2,1],[1,3,1,2,2,2],[1,3,2,1,2,2],[1,3,2,2,1,2],[1,3,2,2,2,1],[2,1,1,2,2,3],[2,1,1,2,3,2],[2,1,1,3,2,2],[2,1,2,1,2,3],[2,1,2,1,3,2],[2,1,2,2,1,3],[2,1,2,2,3,1],[2,1,2,3,1,2],[2,1,2,3,2,1],[2,1,3,1,2,2],[2,1,3,2,1,2],[2,1,3,2,2,1],[2,2,1,1,2,3],[2,2,1,1,3,2],[2,2,1,2,1,3],[2,2,1,2,3,1],[2,2,1,3,1,2],[2,2,1,3,2,1],[2,2,2,1,1,3],[2,2,2,1,3,1],[2,2,2,3,1,1],[2,2,3,1,1,2],[2,2,3,1,2,1],[2,2,3,2,1,1],[2,3,1,1,2,2],[2,3,1,2,1,2],[2,3,1,2,2,1],[2,3,2,1,1,2],[2,3,2,1,2,1],[2,3,2,2,1,1],[3,1,1,2,2,2],[3,1,2,1,2,2],[3,1,2,2,1,2],[3,1,2,2,2,1],[3,2,1,1,2,2],[3,2,1,2,1,2],[3,2,1,2,2,1],[3,2,2,1,1,2],[3,2,2,1,2,1],[3,2,2,2,1,1]]
["AABBBC","AABBCB","AABCBB","AACBBB","ABABBC","ABABCB","ABACBB","ABBABC","ABBACB","ABBBAC","ABBBCA","ABBCAB","ABBCBA","ABCABB","ABCBAB","ABCBBA","ACABBB","ACBABB","ACBBAB","ACBBBA","BAABBC","BAABCB","BAACBB","BABABC","BABACB","BABBAC","BABBCA","BABCAB","BABCBA","BACABB","BACBAB","BACBBA","BBAABC","BBAACB","BBABAC","BBABCA","BBACAB","BBACBA","BBBAAC","BBBACA","BBBCAA","BBCAAB","BBCABA","BBCBAA","BCAABB","BCABAB","BCABBA","BCBAAB","BCBABA","BCBBAA","CAABBB","CABABB","CABBAB","CABBBA","CBAABB","CBABAB","CBABBA","CBBAAB","CBBABA","CBBBAA"]

jq

Works with: jq

Works with gojq, the Go implementation of jq

First, we present the filters for generating the distinct permutations, and then some functions for pretty printing a solution for the task.

<lang jq># Given an array of counts of the nonnegative integers, produce an array reflecting the multiplicities:

  1. e.g. [3,1] => [0,0,0,1]

def items: 

 . as $in
 | reduce range(0;length) as $i ([]; . + [range(0;$in[$i])|$i]);
  1. distinct permutations of the input array, via insertion

def distinct_permutations: 

 # Given an array as input, generate a stream by inserting $x at different positions to the left
 def insert($x):
    ((index([$x]) // length) + 1) as $ix
    | range(0; $ix) as $pos
    | .[0:$pos] + [$x] + .[$pos:];

 if length <= 1 then .
 else
   .[0] as $first
   | .[1:] | distinct_permutations | insert($first)
 end;</lang>

For pretty-printing the results: <lang jq># Input: an array

  1. Output: a stream of arrays

def nwise($n): 

 def w: if length <= $n then . else .[:$n], (.[$n:]|w) end;
 w;

def to_table: 

 nwise(10) | join("  ");</lang>

The task: <lang jq>[2,3,1] | items | [distinct_permutations | join("")] | to_table</lang>

Output:
001112  010112  100112  011012  101012  110012  011102  101102  110102  111002
011120  101120  110120  111020  111200  001121  010121  100121  011021  101021
110021  011201  101201  110201  112001  011210  101210  110210  112010  112100
001211  010211  100211  012011  102011  120011  012101  102101  120101  121001
012110  102110  120110  121010  121100  002111  020111  200111  021011  201011
210011  021101  201101  210101  211001  021110  201110  210110  211010  211100

Julia

With the Combinatorics package, create all permutations and filter out the duplicates

<lang julia>using Combinatorics

catlist(spec) = mapreduce(i -> repeat([i], spec[i]), vcat, 1:length(spec))

alphastringfromintvector(v) = String([Char(Int('A') + i - 1) for i in v])

function testpermwithident(spec) 

   println("\nTesting $spec yielding:")
   for (i, p) in enumerate(unique(collect(permutations(catlist(spec)))))
       print(alphastringfromintvector(p), "  ", i % 10 == 0 ? "\n" : "")
   end

end

testpermwithident([2, 3, 1])

</lang>

Output:
Testing [2, 3, 1] yielding:
AABBBC  AABBCB  AABCBB  AACBBB  ABABBC  ABABCB  ABACBB  ABBABC  ABBACB  ABBBAC
ABBBCA  ABBCAB  ABBCBA  ABCABB  ABCBAB  ABCBBA  ACABBB  ACBABB  ACBBAB  ACBBBA
BAABBC  BAABCB  BAACBB  BABABC  BABACB  BABBAC  BABBCA  BABCAB  BABCBA  BACABB
BACBAB  BACBBA  BBAABC  BBAACB  BBABAC  BBABCA  BBACAB  BBACBA  BBBAAC  BBBACA
BBBCAA  BBCAAB  BBCABA  BBCBAA  BCAABB  BCABAB  BCABBA  BCBAAB  BCBABA  BCBBAA
CAABBB  CABABB  CABBAB  CABBBA  CBAABB  CBABAB  CBABBA  CBBAAB  CBBABA  CBBBAA

Generate directly

Translation of: Tailspin

<lang julia> alpha(s, v) = map(i -> s[i], v)

function distinctPerms(spec) 

   function perm(elements)
       if length(elements) == 1
           deepcopy(elements)
       else
         [pushfirst!(p, elements[k][1]) for k in 1:length(elements) for p in perm(filter(dups -> length(dups) != 0,
           [ if i == k dups[2:end] else dups end
             for (i, dups) in enumerate(elements)]))
         ]
       end
   end
   elements = [fill(x...) for x in enumerate(spec)]
   perm(elements)

end

println(map(p -> join(alpha("ABC", p), ""), distinctPerms([2, 3, 1]))) </lang>

Output:
["AABBBC", "AABBCB", "AABCBB", "AACBBB", "ABABBC", "ABABCB", "ABACBB", "ABBABC", "ABBACB", "ABBBAC", "ABBBCA", "ABBCAB", "ABBCBA", "ABCABB", "ABCBAB", "ABCBBA", "ACABBB", "ACBABB", "ACBBAB", "ACBBBA", "BAABBC", "BAABCB", "BAACBB", "BABABC", "BABACB", "BABBAC", "BABBCA", "BABCAB", "BABCBA", "BACABB", "BACBAB", "BACBBA", "BBAABC", "BBAACB", "BBABAC", "BBABCA", "BBACAB", "BBACBA", "BBBAAC", "BBBACA", "BBBCAA", "BBCAAB", "BBCABA", "BBCBAA", "BCAABB", "BCABAB", "BCABBA", "BCBAAB", "BCBABA", "BCBBAA", "CAABBB", "CABABB", "CABBAB", "CABBBA", "CBAABB", "CBABAB", "CBABBA", "CBBAAB", "CBBABA", "CBBBAA"]

Mathematica/Wolfram Language

<lang Mathematica>ClearAll[PermsFromFrequencies] PermsFromFrequencies[l_List] := Module[{digs}, 

 digs = Flatten[MapIndexed[ReverseApplied[ConstantArray], l]];
 Permutations[digs]
 ]

PermsFromFrequencies[{2, 3, 1}] // Column</lang>

Output:
{1,1,2,2,2,3}
{1,1,2,2,3,2}
{1,1,2,3,2,2}
{1,1,3,2,2,2}
{1,2,1,2,2,3}
{1,2,1,2,3,2}
{1,2,1,3,2,2}
{1,2,2,1,2,3}
{1,2,2,1,3,2}
{1,2,2,2,1,3}
{1,2,2,2,3,1}
{1,2,2,3,1,2}
{1,2,2,3,2,1}
{1,2,3,1,2,2}
{1,2,3,2,1,2}
{1,2,3,2,2,1}
{1,3,1,2,2,2}
{1,3,2,1,2,2}
{1,3,2,2,1,2}
{1,3,2,2,2,1}
{2,1,1,2,2,3}
{2,1,1,2,3,2}
{2,1,1,3,2,2}
{2,1,2,1,2,3}
{2,1,2,1,3,2}
{2,1,2,2,1,3}
{2,1,2,2,3,1}
{2,1,2,3,1,2}
{2,1,2,3,2,1}
{2,1,3,1,2,2}
{2,1,3,2,1,2}
{2,1,3,2,2,1}
{2,2,1,1,2,3}
{2,2,1,1,3,2}
{2,2,1,2,1,3}
{2,2,1,2,3,1}
{2,2,1,3,1,2}
{2,2,1,3,2,1}
{2,2,2,1,1,3}
{2,2,2,1,3,1}
{2,2,2,3,1,1}
{2,2,3,1,1,2}
{2,2,3,1,2,1}
{2,2,3,2,1,1}
{2,3,1,1,2,2}
{2,3,1,2,1,2}
{2,3,1,2,2,1}
{2,3,2,1,1,2}
{2,3,2,1,2,1}
{2,3,2,2,1,1}
{3,1,1,2,2,2}
{3,1,2,1,2,2}
{3,1,2,2,1,2}
{3,1,2,2,2,1}
{3,2,1,1,2,2}
{3,2,1,2,1,2}
{3,2,1,2,2,1}
{3,2,2,1,1,2}
{3,2,2,1,2,1}
{3,2,2,2,1,1}

MiniZinc

<lang MiniZinc> %Permutations with some identical elements. Nigel Galloway, September 9th., 2019 include "count.mzn"; enum N={A,B,C}; array [1..6] of var N: G; constraint count(G,A,2) /\ count(G,C,1); output [show(G)] </lang>

Output:

minizinc --all-solutions produces: 

[C, B, B, B, A, A]
----------
[B, C, B, B, A, A]
----------
[B, B, C, B, A, A]
----------
[B, B, B, C, A, A]
----------
[C, B, B, A, A, B]
----------
[B, C, B, A, A, B]
----------
[B, B, C, A, A, B]
----------
[B, B, B, A, A, C]
----------
[C, B, A, B, A, B]
----------
[B, C, A, B, A, B]
----------
[B, B, A, C, A, B]
----------
[B, B, A, B, A, C]
----------
[C, A, B, B, A, B]
----------
[B, A, C, B, A, B]
----------
[B, A, B, C, A, B]
----------
[B, A, B, B, A, C]
----------
[A, C, B, B, A, B]
----------
[A, B, C, B, A, B]
----------
[A, B, B, C, A, B]
----------
[A, B, B, B, A, C]
----------
[C, B, B, A, B, A]
----------
[B, C, B, A, B, A]
----------
[B, B, C, A, B, A]
----------
[B, B, B, A, C, A]
----------
[C, B, A, A, B, B]
----------
[B, C, A, A, B, B]
----------
[B, B, A, A, C, B]
----------
[B, B, A, A, B, C]
----------
[C, A, B, A, B, B]
----------
[B, A, C, A, B, B]
----------
[B, A, B, A, C, B]
----------
[B, A, B, A, B, C]
----------
[A, C, B, A, B, B]
----------
[A, B, C, A, B, B]
----------
[A, B, B, A, C, B]
----------
[A, B, B, A, B, C]
----------
[C, B, A, B, B, A]
----------
[B, C, A, B, B, A]
----------
[B, B, A, C, B, A]
----------
[B, B, A, B, C, A]
----------
[C, A, A, B, B, B]
----------
[B, A, A, C, B, B]
----------
[B, A, A, B, C, B]
----------
[B, A, A, B, B, C]
----------
[A, C, A, B, B, B]
----------
[A, B, A, C, B, B]
----------
[A, B, A, B, C, B]
----------
[A, B, A, B, B, C]
----------
[C, A, B, B, B, A]
----------
[B, A, C, B, B, A]
----------
[B, A, B, C, B, A]
----------
[B, A, B, B, C, A]
----------
[A, C, B, B, B, A]
----------
[A, B, C, B, B, A]
----------
[A, B, B, C, B, A]
----------
[A, B, B, B, C, A]
----------
[A, A, C, B, B, B]
----------
[A, A, B, C, B, B]
----------
[A, A, B, B, C, B]
----------
[A, A, B, B, B, C]
----------
==========

Nim

Translation of: Go

<lang Nim>import strutils

func shouldSwap(s: string; start, curr: int): bool = 

 for i in start..<curr:
   if s[i] == s[curr]: return false
 return true


func findPerms(s: string; index, n: int; res: var seq[string]) = 

 if index >= n:
   res.add s
   return
 var s = s
 for i in index..<n:
   if s.shouldSwap(index, i):
     swap s[index], s[i]
     findPerms(s, index+1, n, res)
     swap s[index], s[i]


func createSlice(nums: openArray[int]; charSet: string): string = 

 for i, n in nums:
     for _ in 1..n:
       result.add charSet[i]


when isMainModule: 

 var res1, res2, res3: seq[string]
 var nums = @[2, 1]

 var s = createSlice(nums, "12")
 s.findPerms(0, s.len, res1)
 echo res1.join(" ")
 echo()

 nums = @[2, 3, 1]
 s = createSlice(nums, "123")
 findPerms(s, 0, s.len, res2)
 for i, val in res2:
   stdout.write val, if (i + 1) mod 10 == 0: '\n' else: ' '
 echo()

 s = createSlice(nums, "ABC")
 findPerms(s, 0, s.len, res3)
 for i, val in res3:
   stdout.write val, if (i + 1) mod 10 == 0: '\n' else: ' '</lang>
Output:
112 121 211

112223 112232 112322 113222 121223 121232 121322 122123 122132 122213
122231 122321 122312 123221 123212 123122 132221 132212 132122 131222
211223 211232 211322 212123 212132 212213 212231 212321 212312 213221
213212 213122 221123 221132 221213 221231 221321 221312 222113 222131
222311 223121 223112 223211 231221 231212 231122 232121 232112 232211
312221 312212 312122 311222 321221 321212 321122 322121 322112 322211

AABBBC AABBCB AABCBB AACBBB ABABBC ABABCB ABACBB ABBABC ABBACB ABBBAC
ABBBCA ABBCBA ABBCAB ABCBBA ABCBAB ABCABB ACBBBA ACBBAB ACBABB ACABBB
BAABBC BAABCB BAACBB BABABC BABACB BABBAC BABBCA BABCBA BABCAB BACBBA
BACBAB BACABB BBAABC BBAACB BBABAC BBABCA BBACBA BBACAB BBBAAC BBBACA
BBBCAA BBCABA BBCAAB BBCBAA BCABBA BCABAB BCAABB BCBABA BCBAAB BCBBAA
CABBBA CABBAB CABABB CAABBB CBABBA CBABAB CBAABB CBBABA CBBAAB CBBBAA

Pascal

modified version to get permutations of different lenght.
Most time consuming is appending to the stringlist.So I limited that to 10000
One can use the string directly in EvaluatePerm. <lang pascal>program PermWithRep;//of different length {$IFDEF FPC} 

 {$mode Delphi}
 {$Optimization ON,All}

{$ELSE} 

 {$APPTYPE CONSOLE}

{$ENDIF} uses 

 sysutils,classes {for stringlist};

const 

 cTotalSum = 16;
 cMaxCardsOnDeck = cTotalSum;
 CMaxCardsUsed   = cTotalSum;

type 

 tDeckIndex     = 0..cMaxCardsOnDeck-1;
 tSequenceIndex = 0..CMaxCardsUsed;
 tDiffCardCount = Byte;//A..Z

 tSetElem     = packed record
                  Elemcount : tDeckIndex;
                  Elem  : tDiffCardCount;
                end;

 tRemSet = array [low(tDeckIndex)..High(tDeckIndex)] of tSetElem;
 tpRemSet = ^tRemSet;
 tRemainSet      = array [low(tSequenceIndex)..High(tSequenceIndex)] of tRemSet;
 tCardSequence   = array [low(tSequenceIndex)..High(tSequenceIndex)] of tDiffCardCount;

var {$ALIGN 32} 

 RemainSets     : tRemainSet;
 CardString    : AnsiString;
 CS : pchar;
 sl :TStringList;
 gblMaxCardsUsed,
 gblMaxUsedIdx,
 gblPermCount : NativeInt;

//***************************************************************************** procedure Out_SL(const sl:TStringlIst;colCnt:NativeInt); var 

 j,i : NativeInt;

begin 

 j := colCnt;
 For i := 0 to sl.count-1 do
 Begin
   write(sl[i],' ');
   dec(j);
   if j= 0 then
   Begin
     writeln;
     j := colCnt;
   end;
 end;
 if j <> colCnt then
   writeln;

end;

procedure SetClear(var ioSet:tRemSet); begin 

 fillChar(ioSet[0],SizeOf(ioSet),#0);

end;

procedure SetInit(var ioSet:tRemSet;const inSet:tRemSet); var 

 i,j,k,sum : integer;

begin 

 ioSet := inSet;
 sum := 0;
 k := 0;
 write('Initial set : ');
 For i := Low(ioSet) to High(ioSet) do
 Begin
   j := inSet[i].ElemCount;
   if j <> 0 then
     inc(k);
   sum += j;
   For j := j downto 1 do
     write(chr(inSet[i].Elem));
 end;
 gblMaxCardsUsed := sum;
 gblMaxUsedIdx := k;
 writeln(' lenght: ',sum,' different elements : ',k);

end;

procedure EvaluatePerm; Begin 

 //append maximal 10000 strings
 if gblPermCount < 10000 then
   sl.append(CS);

end;

procedure Permute(depth,MaxCardsUsed:NativeInt); var 

 pSetElem : tpRemSet;//^tSetElem;
 i : NativeInt;

begin 

 i := 0;
 pSetElem := @RemainSets[depth];
 repeat
   if pSetElem^[i].Elemcount > 0 then begin
     //take one of the same elements of the stack
     //insert in result here string
     CS[depth] := chr(pSetElem^[i].Elem);
     //done one permutation
     IF depth = MaxCardsUsed then
     begin
       inc(gblpermCount);
       EvaluatePerm;
     end
     else
     begin
       RemainSets[depth+1]:=RemainSets[depth];
       //remove one element
       dec(RemainSets[depth+1][i].ElemCount);
       Permute(depth+1,MaxCardsUsed);
     end;
   end;
   //move on to the next Elem
   inc(i);
 until i >= gblMaxUsedIdx;

end;

procedure Permutate(MaxCardsUsed:NativeInt); Begin 

 gblpermCount := 0;
 if MaxCardsUsed > gblMaxCardsUsed then
   MaxCardsUsed := gblMaxCardsUsed;

 if MaxCardsUsed>0 then
 Begin
   setlength(CardString,MaxCardsUsed);
   CS := @CardString[1];
   permute(0,MaxCardsUsed-1)
 end;

end;

var 

 Manifolds : tRemSet;
 j :nativeInt;

Begin 

 SetClear(Manifolds);
 with Manifolds[0] do
 begin
   Elemcount := 2; Elem := Ord('A');
 end;
 with Manifolds[1] do
 begin
   Elemcount := 3; Elem := Ord('B');
 end;
 with Manifolds[2] do
 begin
   Elemcount := 1; Elem := Ord('C');
 end;

 try
   sl := TStringList.create;

   SetInit(RemainSets[0], Manifolds);
   j := gblMaxCardsUsed;
   writeln('Count of elements: ',j);
   while j > 1 do
   begin
     sl.clear;
     Permutate(j);
     writeln('Length ',j:2,' Permutations ',gblpermCount:7);
     Out_SL(sl,80 DIV (Length(CS)+1));
     writeln;
     dec(j);
   end;
   //change to 1,2,3
   Manifolds[0].Elem := Ord('1');
   Manifolds[1].Elem := Ord('2');
   Manifolds[2].Elem := Ord('3');

   SetInit(RemainSets[0], Manifolds);
   j := gblMaxCardsUsed;
   writeln('Count of elements: ',j);
   while j > 1 do
   begin
     sl.clear;
     Permutate(j);
     writeln('Length ',j:2,' Permutations ',gblpermCount:7);
     Out_SL(sl,80 DIV (Length(CS)+1));
     writeln;
     dec(j);
   end;

 //extend by 3 more elements
 with Manifolds[3] do
 begin
   Elemcount := 2;   Elem := Ord('4');
 end;
 with Manifolds[4] do
 begin
   Elemcount := 3;  Elem := Ord('5');
 end;
 with Manifolds[5] do
 begin
   Elemcount := 1;  Elem := Ord('6');
 end;
 SetInit(RemainSets[0], Manifolds);
 j := gblMaxCardsUsed;
 writeln('Count of elements: ',j);
 sl.clear;
 Permutate(j);
 writeln('Length ',j:2,' Permutations ',gblpermCount:7);
 //Out_SL(sl,80 DIV (Length(CS)+1));
 writeln;

 except
   writeln(' Stringlist Error ');
 end;
 sl.free;

end.</lang>

Output:
Initial set : AABBBC lenght: 6
Count of elements: 6
Length  6 Permutations      60
AABBBC AABBCB AABCBB AACBBB ABABBC ABABCB ABACBB ABBABC ABBACB ABBBAC ABBBCA
ABBCAB ABBCBA ABCABB ABCBAB ABCBBA ACABBB ACBABB ACBBAB ACBBBA BAABBC BAABCB
BAACBB BABABC BABACB BABBAC BABBCA BABCAB BABCBA BACABB BACBAB BACBBA BBAABC
BBAACB BBABAC BBABCA BBACAB BBACBA BBBAAC BBBACA BBBCAA BBCAAB BBCABA BBCBAA
BCAABB BCABAB BCABBA BCBAAB BCBABA BCBBAA CAABBB CABABB CABBAB CABBBA CBAABB
CBABAB CBABBA CBBAAB CBBABA CBBBAA

Length  5 Permutations      60
AABBB AABBC AABCB AACBB ABABB ABABC ABACB ABBAB ABBAC ABBBA ABBBC ABBCA ABBCB
ABCAB ABCBA ABCBB ACABB ACBAB ACBBA ACBBB BAABB BAABC BAACB BABAB BABAC BABBA
BABBC BABCA BABCB BACAB BACBA BACBB BBAAB BBAAC BBABA BBABC BBACA BBACB BBBAA
BBBAC BBBCA BBCAA BBCAB BBCBA BCAAB BCABA BCABB BCBAA BCBAB BCBBA CAABB CABAB
CABBA CABBB CBAAB CBABA CBABB CBBAA CBBAB CBBBA

Length  4 Permutations      38
AABB AABC AACB ABAB ABAC ABBA ABBB ABBC ABCA ABCB ACAB ACBA ACBB BAAB BAAC BABA
BABB BABC BACA BACB BBAA BBAB BBAC BBBA BBBC BBCA BBCB BCAA BCAB BCBA BCBB CAAB
CABA CABB CBAA CBAB CBBA CBBB

Length  3 Permutations      19
AAB AAC ABA ABB ABC ACA ACB BAA BAB BAC BBA BBB BBC BCA BCB CAA CAB CBA CBB

Length  2 Permutations       8
AA AB AC BA BB BC CA CB

Initial set : 112223 lenght: 6
Count of elements: 6
Length  6 Permutations      60
112223 112232 112322 113222 121223 121232 121322 122123 122132 122213 122231
122312 122321 123122 123212 123221 131222 132122 132212 132221 211223 211232
211322 212123 212132 212213 212231 212312 212321 213122 213212 213221 221123
221132 221213 221231 221312 221321 222113 222131 222311 223112 223121 223211
231122 231212 231221 232112 232121 232211 311222 312122 312212 312221 321122
321212 321221 322112 322121 322211

Length  5 Permutations      60
11222 11223 11232 11322 12122 12123 12132 12212 12213 12221 12223 12231 12232
12312 12321 12322 13122 13212 13221 13222 21122 21123 21132 21212 21213 21221
21223 21231 21232 21312 21321 21322 22112 22113 22121 22123 22131 22132 22211
22213 22231 22311 22312 22321 23112 23121 23122 23211 23212 23221 31122 31212
31221 31222 32112 32121 32122 32211 32212 32221

Length  4 Permutations      38
1122 1123 1132 1212 1213 1221 1222 1223 1231 1232 1312 1321 1322 2112 2113 2121
2122 2123 2131 2132 2211 2212 2213 2221 2223 2231 2232 2311 2312 2321 2322 3112
3121 3122 3211 3212 3221 3222

Length  3 Permutations      19
112 113 121 122 123 131 132 211 212 213 221 222 223 231 232 311 312 321 322

Length  2 Permutations       8
11 12 13 21 22 23 31 32

Initial set : 112223445556 lenght: 12
Count of elements: 12
Length 12 Permutations 3326400

Perl

Library: ntheory

<lang perl>use ntheory qw<formultiperm>;

formultiperm { print join(,@_) . ' ' } [<1 1 2>]; print "\n\n"; formultiperm { print join(,@_) . ' ' } [<1 1 2 2 2 3>]; print "\n\n"; formultiperm { print join(,@_) . ' ' } [split //,'AABBBC']; print "\n";</lang>

Output:
112 121 211

112223 112232 112322 113222 121223 121232 121322 122123 122132 122213 122231 122312 122321 123122 123212 123221 131222 132122 132212 132221 211223 211232 211322 212123 212132 212213 212231 212312 212321 213122 213212 213221 221123 221132 221213 221231 221312 221321 222113 222131 222311 223112 223121 223211 231122 231212 231221 232112 232121 232211 311222 312122 312212 312221 321122 321212 321221 322112 322121 322211

AABBBC AABBCB AABCBB AACBBB ABABBC ABABCB ABACBB ABBABC ABBACB ABBBAC ABBBCA ABBCAB ABBCBA ABCABB ABCBAB ABCBBA ACABBB ACBABB ACBBAB ACBBBA BAABBC BAABCB BAACBB BABABC BABACB BABBAC BABBCA BABCAB BABCBA BACABB BACBAB BACBBA BBAABC BBAACB BBABAC BBABCA BBACAB BBACBA BBBAAC BBBACA BBBCAA BBCAAB BBCABA BBCBAA BCAABB BCABAB BCABBA BCBAAB BCBABA BCBBAA CAABBB CABABB CABBAB CABBBA CBAABB CBABAB CBABBA CBBAAB CBBABA CBBBAA

Phix

Translation of: Go
with javascript_semantics 
function shouldSwap(string s, integer start, curr)
    for i=start to curr-1 do
        if s[i] == s[curr] then
            return false
        end if
    end for
    return true
end function
 
function findPerms(string s, integer i=1, sequence res={})
    if i>length(s) then
        res = append(res, s)
    else
        for j=i to length(s) do
            if shouldSwap(s, i, j) then
                integer {si, sj} = {s[i], s[j]}
                s[i] = sj
                s[j] = si
                res = findPerms(s, i+1, res)
                s[i] = si
                s[j] = sj
            end if
        end for
    end if
    return res
end function
 
function createSlice(sequence nums, string charSet)
    string chars = ""
    for i=1 to length(nums) do
        chars &= repeat(charSet[i],nums[i])
    end for
    return chars
end function
 
pp(findPerms(createSlice({2,1}, "12")))     -- (=== findPerms("112"))
pp(findPerms(createSlice({2,3,1}, "123")))  -- (=== findPerms("112223"))
pp(findPerms(createSlice({2,3,1}, "ABC")))  -- (=== findPerms("AABBBC"))
Output:
{`112`, `121`, `211`}
{`112223`, `112232`, `112322`, `113222`, `121223`, `121232`, `121322`,
 `122123`, `122132`, `122213`, `122231`, `122321`, `122312`, `123221`,
 `123212`, `123122`, `132221`, `132212`, `132122`, `131222`, `211223`,
 `211232`, `211322`, `212123`, `212132`, `212213`, `212231`, `212321`,
 `212312`, `213221`, `213212`, `213122`, `221123`, `221132`, `221213`,
 `221231`, `221321`, `221312`, `222113`, `222131`, `222311`, `223121`,
 `223112`, `223211`, `231221`, `231212`, `231122`, `232121`, `232112`,
 `232211`, `312221`, `312212`, `312122`, `311222`, `321221`, `321212`,
 `321122`, `322121`, `322112`, `322211`}
{`AABBBC`, `AABBCB`, `AABCBB`, `AACBBB`, `ABABBC`, `ABABCB`, `ABACBB`,
 `ABBABC`, `ABBACB`, `ABBBAC`, `ABBBCA`, `ABBCBA`, `ABBCAB`, `ABCBBA`,
 `ABCBAB`, `ABCABB`, `ACBBBA`, `ACBBAB`, `ACBABB`, `ACABBB`, `BAABBC`,
 `BAABCB`, `BAACBB`, `BABABC`, `BABACB`, `BABBAC`, `BABBCA`, `BABCBA`,
 `BABCAB`, `BACBBA`, `BACBAB`, `BACABB`, `BBAABC`, `BBAACB`, `BBABAC`,
 `BBABCA`, `BBACBA`, `BBACAB`, `BBBAAC`, `BBBACA`, `BBBCAA`, `BBCABA`,
 `BBCAAB`, `BBCBAA`, `BCABBA`, `BCABAB`, `BCAABB`, `BCBABA`, `BCBAAB`,
 `BCBBAA`, `CABBBA`, `CABBAB`, `CABABB`, `CAABBB`, `CBABBA`, `CBABAB`,
 `CBAABB`, `CBBABA`, `CBBAAB`, `CBBBAA`}

You can also use a builtin function to produce exactly the same output (createSlice omitted for clarity) 

with javascript_semantics 
requires("1.0.2")
pp(permutes("112"))
pp(permutes("112223"))
pp(permutes("AABBBC"))

Python

Set filters out unique permutations <lang python>

  1. Aamrun, 5th October 2021

from itertools import permutations

numList = [2,3,1]

baseList = []

for i in numList: 

   for j in range(0,i):
       baseList.append(i)

stringDict = {'A':2,'B':3,'C':1}

baseString=""

for i in stringDict: 

   for j in range(0,stringDict[i]):
       baseString+=i

print("Permutations for " + str(baseList) + " : ") [print(i) for i in set(permutations(baseList))]

print("Permutations for " + baseString + " : ") [print(i) for i in set(permutations(baseString))] </lang>

Output:
Permutations for [2, 2, 3, 3, 3, 1] : 
(2, 1, 3, 2, 3, 3)
(3, 2, 3, 3, 1, 2)
(3, 2, 3, 1, 2, 3)
(3, 3, 1, 2, 2, 3)
(3, 3, 2, 2, 3, 1)
(3, 1, 3, 3, 2, 2)
(3, 1, 3, 2, 3, 2)
(2, 3, 3, 3, 1, 2)
(3, 3, 1, 3, 2, 2)
(3, 1, 2, 3, 2, 3)
(3, 2, 2, 1, 3, 3)
(3, 2, 3, 2, 1, 3)
(3, 1, 3, 2, 2, 3)
(2, 3, 1, 3, 2, 3)
(3, 2, 1, 2, 3, 3)
(2, 3, 3, 2, 1, 3)
(2, 3, 1, 3, 3, 2)
(3, 3, 2, 3, 2, 1)
(3, 3, 2, 2, 1, 3)
(2, 2, 3, 3, 3, 1)
(2, 3, 1, 2, 3, 3)
(3, 3, 2, 3, 1, 2)
(3, 3, 3, 2, 2, 1)
(2, 1, 2, 3, 3, 3)
(2, 3, 2, 3, 3, 1)
(2, 1, 3, 3, 3, 2)
(2, 2, 3, 3, 1, 3)
(3, 1, 2, 2, 3, 3)
(2, 3, 2, 1, 3, 3)
(3, 2, 1, 3, 3, 2)
(1, 3, 3, 3, 2, 2)
(3, 3, 3, 2, 1, 2)
(2, 3, 2, 3, 1, 3)
(3, 2, 2, 3, 3, 1)
(1, 3, 2, 2, 3, 3)
(2, 1, 3, 3, 2, 3)
(3, 2, 1, 3, 2, 3)
(1, 3, 3, 2, 2, 3)
(1, 3, 3, 2, 3, 2)
(1, 2, 3, 3, 3, 2)
(2, 3, 3, 1, 3, 2)
(3, 3, 2, 1, 3, 2)
(1, 2, 3, 2, 3, 3)
(3, 3, 2, 1, 2, 3)
(3, 2, 3, 3, 2, 1)
(1, 3, 2, 3, 3, 2)
(1, 2, 3, 3, 2, 3)
(2, 3, 3, 1, 2, 3)
(3, 2, 2, 3, 1, 3)
(1, 3, 2, 3, 2, 3)
(3, 1, 2, 3, 3, 2)
(2, 3, 3, 2, 3, 1)
(3, 3, 1, 2, 3, 2)
(2, 2, 3, 1, 3, 3)
(3, 3, 3, 1, 2, 2)
(1, 2, 2, 3, 3, 3)
(3, 2, 3, 1, 3, 2)
(2, 2, 1, 3, 3, 3)
(2, 3, 3, 3, 2, 1)
(3, 2, 3, 2, 3, 1)
Permutations for AABBBC : 
('B', 'B', 'A', 'C', 'B', 'A')
('B', 'C', 'B', 'A', 'B', 'A')
('A', 'B', 'A', 'B', 'B', 'C')
('B', 'A', 'B', 'B', 'C', 'A')
('C', 'B', 'A', 'B', 'B', 'A')
('A', 'A', 'B', 'C', 'B', 'B')
('A', 'B', 'B', 'C', 'A', 'B')
('B', 'C', 'B', 'B', 'A', 'A')
('B', 'B', 'A', 'A', 'C', 'B')
('A', 'B', 'C', 'B', 'B', 'A')
('A', 'C', 'A', 'B', 'B', 'B')
('B', 'B', 'B', 'A', 'A', 'C')
('C', 'B', 'B', 'A', 'B', 'A')
('A', 'A', 'C', 'B', 'B', 'B')
('C', 'A', 'A', 'B', 'B', 'B')
('B', 'C', 'A', 'A', 'B', 'B')
('A', 'B', 'B', 'A', 'C', 'B')
('B', 'B', 'C', 'A', 'B', 'A')
('A', 'A', 'B', 'B', 'B', 'C')
('C', 'B', 'B', 'B', 'A', 'A')
('B', 'C', 'A', 'B', 'B', 'A')
('C', 'A', 'B', 'A', 'B', 'B')
('B', 'A', 'A', 'B', 'C', 'B')
('B', 'B', 'B', 'C', 'A', 'A')
('B', 'B', 'C', 'B', 'A', 'A')
('B', 'A', 'B', 'C', 'A', 'B')
('A', 'C', 'B', 'B', 'B', 'A')
('B', 'B', 'A', 'A', 'B', 'C')
('C', 'B', 'A', 'A', 'B', 'B')
('C', 'A', 'B', 'B', 'A', 'B')
('B', 'A', 'A', 'C', 'B', 'B')
('B', 'C', 'B', 'A', 'A', 'B')
('B', 'B', 'A', 'B', 'A', 'C')
('A', 'B', 'B', 'A', 'B', 'C')
('B', 'A', 'A', 'B', 'B', 'C')
('A', 'B', 'B', 'C', 'B', 'A')
('B', 'A', 'C', 'B', 'B', 'A')
('C', 'B', 'A', 'B', 'A', 'B')
('B', 'A', 'B', 'A', 'C', 'B')
('A', 'B', 'A', 'B', 'C', 'B')
('A', 'B', 'C', 'A', 'B', 'B')
('B', 'B', 'B', 'A', 'C', 'A')
('A', 'B', 'B', 'B', 'A', 'C')
('B', 'B', 'A', 'C', 'A', 'B')
('A', 'B', 'C', 'B', 'A', 'B')
('C', 'A', 'B', 'B', 'B', 'A')
('A', 'A', 'B', 'B', 'C', 'B')
('B', 'A', 'B', 'B', 'A', 'C')
('A', 'B', 'A', 'C', 'B', 'B')
('A', 'C', 'B', 'A', 'B', 'B')
('B', 'A', 'C', 'A', 'B', 'B')
('B', 'B', 'A', 'B', 'C', 'A')
('B', 'C', 'A', 'B', 'A', 'B')
('C', 'B', 'B', 'A', 'A', 'B')
('B', 'A', 'B', 'A', 'B', 'C')
('B', 'A', 'B', 'C', 'B', 'A')
('B', 'A', 'C', 'B', 'A', 'B')
('A', 'B', 'B', 'B', 'C', 'A')
('B', 'B', 'C', 'A', 'A', 'B')
('A', 'C', 'B', 'B', 'A', 'B')

Raku

(formerly Perl 6)

Works with: Rakudo version 2019.07

<lang perl6>sub permutations-with-some-identical-elements ( @elements, @reps = () ) { 

   with @elements { (@reps ?? flat $_ Zxx @reps !! flat .keys.map(*+1) Zxx .values).permutations».join.unique }
}

for (<2 1>,), (<2 3 1>,), (<A B C>, <2 3 1>), (<🦋 ⚽ 🙄>, <2 2 1>) { 

   put permutations-with-some-identical-elements |$_;
   say ;

}</lang>

Output:
112 121 211

112223 112232 112322 113222 121223 121232 121322 122123 122132 122213 122231 122312 122321 123122 123212 123221 131222 132122 132212 132221 211223 211232 211322 212123 212132 212213 212231 212312 212321 213122 213212 213221 221123 221132 221213 221231 221312 221321 222113 222131 222311 223112 223121 223211 231122 231212 231221 232112 232121 232211 311222 312122 312212 312221 321122 321212 321221 322112 322121 322211

AABBBC AABBCB AABCBB AACBBB ABABBC ABABCB ABACBB ABBABC ABBACB ABBBAC ABBBCA ABBCAB ABBCBA ABCABB ABCBAB ABCBBA ACABBB ACBABB ACBBAB ACBBBA BAABBC BAABCB BAACBB BABABC BABACB BABBAC BABBCA BABCAB BABCBA BACABB BACBAB BACBBA BBAABC BBAACB BBABAC BBABCA BBACAB BBACBA BBBAAC BBBACA BBBCAA BBCAAB BBCABA BBCBAA BCAABB BCABAB BCABBA BCBAAB BCBABA BCBBAA CAABBB CABABB CABBAB CABBBA CBAABB CBABAB CBABBA CBBAAB CBBABA CBBBAA

🦋🦋⚽⚽🙄 🦋🦋⚽🙄⚽ 🦋🦋🙄⚽⚽ 🦋⚽🦋⚽🙄 🦋⚽🦋🙄⚽ 🦋⚽⚽🦋🙄 🦋⚽⚽🙄🦋 🦋⚽🙄🦋⚽ 🦋⚽🙄⚽🦋 🦋🙄🦋⚽⚽ 🦋🙄⚽🦋⚽ 🦋🙄⚽⚽🦋 ⚽🦋🦋⚽🙄 ⚽🦋🦋🙄⚽ ⚽🦋⚽🦋🙄 ⚽🦋⚽🙄🦋 ⚽🦋🙄🦋⚽ ⚽🦋🙄⚽🦋 ⚽⚽🦋🦋🙄 ⚽⚽🦋🙄🦋 ⚽⚽🙄🦋🦋 ⚽🙄🦋🦋⚽ ⚽🙄🦋⚽🦋 ⚽🙄⚽🦋🦋 🙄🦋🦋⚽⚽ 🙄🦋⚽🦋⚽ 🙄🦋⚽⚽🦋 🙄⚽🦋🦋⚽ 🙄⚽🦋⚽🦋 🙄⚽⚽🦋🦋

REXX

shows permutation list

<lang rexx>/*REXX program computes and displays the permutations with some identical elements. */ parse arg g /*obtain optional arguments from the CL*/ if g= | g="," then g= 2 3 1 /*Not specified? Then use the defaults*/

  1. = words(g) /*obtain the number of source items. */

@= left('ABCDEFGHIJKLMNOPQRSTUVWXYZ', #) /*@: the (output) letters to be used.*/ LO= /*LO: the start of the sequence. */ HI= /*HI: " end " " " */ 

     do i=1  for #;      @.i= word(g, i)        /*get number of characters for an arg. */
     LO= LO || copies(i, @.i)                   /*build the  LO  number for the range. */
     HI=       copies(i, @.i) || HI             /*  "    "   HI     "    "   "    "    */
     end   /*i*/

$= /*initialize the output string to null.*/ 

     do j=LO  to  HI                            /*generate the enumerated output string*/
     if verify(j, LO)\==0  then iterate         /*An invalid digital string?  Then skip*/
        do k=1  for #                           /*parse string for correct # of digits.*/
        if countstr(k, j)\==@.k  then iterate j /*Incorrect number of digits?  Skip.   */
        end   /*k*/
     $= $ j                                     /*append digital string to the list.   */
     end      /*j*/
                                                /*stick a fork in it,  we're all done. */

say 'number of permutations: ' words($) say say strip(translate($, @, left(123456789, #) ) ) /*display the translated string to term*/</lang>

output   when using the inputs of:     2   1 
number of permutations:  3

AAB ABA BAA
output   when using the default inputs:     2   3   1 
number of permutations:  60

AABBBC AABBCB AABCBB AACBBB ABABBC ABABCB ABACBB ABBABC ABBACB ABBBAC ABBBCA ABBCAB ABBCBA ABCABB ABCBAB ABCBBA ACABBB ACBABB ACBBAB ACBBBA BAABBC BAABCB BAACBB BABABC BABACB BABBAC BABBCA BABCAB BABCBA BACABB BACBAB BACBBA BBAABC BBAACB BBABAC BBABCA BBACAB BBACBA BBBAAC BBBACA BBBCAA BBCAAB BBCABA BBCBAA BCAABB BCABAB BCABBA BCBAAB BCBABA BCBBAA CAABBB CABABB CABBAB CABBBA CBAABB CBABAB CBABBA CBBAAB CBBABA CBBBAA

only shows permutation count

If any of the arguments is negative, the list of the permutations is suppressed, only the permutation count is shown. <lang rexx>/*REXX program computes and displays the permutations with some identical elements. */ parse arg g /*obtain optional arguments from the CL*/ if g= | g="," then g= 2 3 1 /*Not specified? Then use the defaults*/

  1. = words(g) /*obtain the number of source items. */

@= left('ABCDEFGHIJKLMNOPQRSTUVWXYZ', #) /*@: the (output) letters to be used.*/ show= 1 /*if = 1, will show permutation list. */ sum= 0 LO= /*LO: the start of the sequence. */ HI= /*HI: " end " " " */ 

     do i=1  for #;        y= word(g, i)        /*get number of characters for an arg. */
     show= show & y>=0;    a= abs(y)            /*Is it negative?  Then don't show list*/
     sum= sum + a;       @.i= a                 /*use the absolute value of an argument*/
     LO= LO || copies(i, @.i)                   /*build the  LO  number for the range. */
     HI=       copies(i, @.i) || HI             /*  "    "   HI     "    "   "    "    */
     end   /*i*/

$= /*initialize the output string to null.*/ numeric digits max(9, sum) /*ensure enough numeric decimal digits.*/ 

     do j=LO  to  HI                            /*generate the enumerated output string*/
     if verify(j, LO)\==0  then iterate         /*An invalid digital string?  Then skip*/
        do k=1  for #                           /*parse string for correct # of digits.*/
        if countstr(k, j)\==@.k  then iterate j /*Incorrect number of digits?  Skip.   */
        end   /*k*/
     $= $ j                                     /*append digital string to the list.   */
     end      /*j*/
                                                /*stick a fork in it,  we're all done. */

say 'number of permutations: ' words($) /*# perms with some identical elements.*/ say if show then say strip(translate($, @, left(123456789, #) ) ) /*display translated str.*/</lang>

output   when using the inputs of:     -2   3   1   1   1 
number of permutations:  3360
output   when using the inputs of:     -1   2   1   2   1 
number of permutations:  1260
output   when using the inputs of:     -1   2   3   2   1 
number of permutations:  15120
output   when using the inputs of:     -1   1   5   2 
number of permutations:  1512
output   when using the inputs of:     -1   1   1   1   6 
number of permutations:  5040

Sidef

Simple implementation, by filtering out the duplicated permutations. <lang ruby>func permutations_with_some_identical_elements (reps) { 

   reps.map_kv {|k,v| v.of(k+1)... }.permutations.uniq

}

say permutations_with_some_identical_elements([2,1]).map{.join}.join(' ') say permutations_with_some_identical_elements([2,3,1]).map{.join}.join(' ')</lang>

Output:
112 121 211
112223 112232 112322 113222 121223 121232 121322 122123 122132 122213 122231 122312 122321 123122 123212 123221 131222 132122 132212 132221 211223 211232 211322 212123 212132 212213 212231 212312 212321 213122 213212 213221 221123 221132 221213 221231 221312 221321 222113 222131 222311 223112 223121 223211 231122 231212 231221 232112 232121 232211 311222 312122 312212 312221 321122 321212 321221 322112 322121 322211

More efficient approach, by generating the permutations without duplicates: <lang ruby>func next_uniq_perm (array) { 

   var k = array.end
   return ([], false) if (k < 0)
   var i = k-1

   while ((i >= 0) && (array[i] >= array[i+1])) {
       --i
   }

   return (array.flip, false) if (i == -1)

   if (array[i+1] > array[k]) {
       array = [array.slice(0, i)..., array.slice(i+1, k).flip...]
   }

   var j = i+1
   while (array[i] >= array[j]) {
       j++
   }

   array.clone!
   array.swap(i,j)

   return (array, true)

}

func unique_permutations(array, reps=[]) { 

   var perm = (reps ? reps : array).map_kv {|k,v| v.of { reps ? array[k] : (k+1) }... }
   var perms = [perm]
   loop {
       (perm, var more) = next_uniq_perm(perm)
       break if !more
       perms << perm
   }
   return perms

}

for a,b in ([[[2,1]], 2,3,1, [%w(A B C), [2,3,1]]]) { 

   say "\nPermutations with array = #{a}#{b ? \" and reps = #{b}\" : }:"
   say unique_permutations(a,b).map{.join}.join(' ')

}</lang>

Output:
Permutations with array = [2, 1]:
112 121 211

Permutations with array = [2, 3, 1]:
112223 112232 112322 113222 121223 121232 121322 122123 122132 122213 122231 122312 122321 123122 123212 123221 131222 132122 132212 132221 211223 211232 211322 212123 212132 212213 212231 212312 212321 213122 213212 213221 221123 221132 221213 221231 221312 221321 222113 222131 222311 223112 223121 223211 231122 231212 231221 232112 232121 232211 311222 312122 312212 312221 321122 321212 321221 322112 322121 322211

Permutations with array = ["A", "B", "C"] and reps = [2, 3, 1]:
AABBBC AABBCB AABCBB AACBBB ABABBC ABABCB ABACBB ABBABC ABBACB ABBBAC ABBBCA ABBCAB ABBCBA ABCABB ABCBAB ABCBBA ACABBB ACBABB ACBBAB ACBBBA BAABBC BAABCB BAACBB BABABC BABACB BABBAC BABBCA BABCAB BABCBA BACABB BACBAB BACBBA BBAABC BBAACB BBABAC BBABCA BBACAB BBACBA BBBAAC BBBACA BBBCAA BBCAAB BBCABA BBCBAA BCAABB BCABAB BCABBA BCBAAB BCBABA BCBBAA CAABBB CABABB CABBAB CABBBA CBAABB CBABAB CBABBA CBBAAB CBBABA CBBBAA

Tailspin

Creates lots of new arrays, which might be wasteful. <lang tailspin> templates distinctPerms 

 templates perms
   when <[](1)> do $(1) !
   otherwise
     def elements: $;
     1..$::length -> \(
       def k: $;
       def tail: $elements -> \[i](
         when <?($i <=$k>)> do $ -> \(<[](2..)> $(2..last)!\) !
         otherwise $!
       \);
       $tail -> perms -> [$elements($k;1), $...] !
     \) !
 end perms
 $ -> \[i]([1..$ -> $i] !\) -> perms !

end distinctPerms

def alpha: ['ABC'...]; [[2,3,1] -> distinctPerms -> '$alpha($)...;' ] -> !OUT::write </lang>

Output:
[AABBBC, AABBCB, AABCBB, AACBBB, ABABBC, ABABCB, ABACBB, ABBABC, ABBACB, ABBBAC, ABBBCA, ABBCAB, ABBCBA, ABCABB, ABCBAB, ABCBBA, ACABBB, ACBABB, ACBBAB, ACBBBA, BAABBC, BAABCB, BAACBB, BABABC, BABACB, BABBAC, BABBCA, BABCAB, BABCBA, BACABB, BACBAB, BACBBA, BBAABC, BBAACB, BBABAC, BBABCA, BBACAB, BBACBA, BBBAAC, BBBACA, BBBCAA, BBCAAB, BBCABA, BBCBAA, BCAABB, BCABAB, BCABBA, BCBAAB, BCBABA, BCBBAA, CAABBB, CABABB, CABBAB, CABBBA, CBAABB, CBABAB, CBABBA, CBBAAB, CBBABA, CBBBAA]

Work in place (slightly modified from the Go solution to preserve lexical order)

Translation of: Go

<lang tailspin> templates distinctPerms 

 templates shouldSwap&{start:}
       when <?($@distinctPerms($start..~$) <~[<=$@distinctPerms($)>]>)> do $!
 end shouldSwap
 templates findPerms
   when <$@distinctPerms::length..> do $@distinctPerms !
   otherwise
     def index: $;
     $index..$@distinctPerms::length -> shouldSwap&{start: $index}
     -> \(
         @findPerms: $;
         @distinctPerms([$, $index]): $@distinctPerms([$index, $])...;
         $index + 1 -> findPerms !
     \) !
     @distinctPerms([$@findPerms, $index..~$@findPerms]): $@distinctPerms($index..$@findPerms)...;
 end findPerms
 @: $ -> \[i](1..$ -> $i !\);
 1 -> findPerms !

end distinctPerms

def alpha: ['ABC'...]; [[2,3,1] -> distinctPerms -> '$alpha($)...;' ] -> !OUT::write </lang>

Output:
[AABBBC, AABBCB, AABCBB, AACBBB, ABABBC, ABABCB, ABACBB, ABBABC, ABBACB, ABBBAC, ABBBCA, ABBCAB, ABBCBA, ABCABB, ABCBAB, ABCBBA, ACABBB, ACBABB, ACBBAB, ACBBBA, BAABBC, BAABCB, BAACBB, BABABC, BABACB, BABBAC, BABBCA, BABCAB, BABCBA, BACABB, BACBAB, BACBBA, BBAABC, BBAACB, BBABAC, BBABCA, BBACAB, BBACBA, BBBAAC, BBBACA, BBBCAA, BBCAAB, BBCABA, BBCBAA, BCAABB, BCABAB, BCABBA, BCBAAB, BCBABA, BCBBAA, CAABBB, CABABB, CABBAB, CABBBA, CBAABB, CBABAB, CBABBA, CBBAAB, CBBABA, CBBBAA]

Wren

Library: Wren-perm

<lang ecmascript>import "./perm" for Perm

var createList = Fn.new { |nums, charSet| 

   var chars = []
   for (i in 0...nums.count) {
       for (j in 0...nums[i]) chars.add(charSet[i])
   }
   return chars

}

var nums = [2, 1] var a = createList.call(nums, "12") System.print(Perm.listDistinct(a).map { |p| p.join() }.toList) System.print()

nums = [2, 3, 1] a = createList.call(nums, "123") System.print(Perm.listDistinct(a).map { |p| p.join() }.toList) System.print()

a = createList.call(nums, "ABC") System.print(Perm.listDistinct(a).map { |p| p.join() }.toList)</lang>

Output:
[112, 121, 211]

[112223, 112232, 112322, 113222, 121223, 121232, 121322, 122123, 122132, 122213, 122231, 122321, 122312, 123221, 123212, 123122, 132221, 132212, 132122, 131222, 211223, 211232, 211322, 212123, 212132, 212213, 212231, 212321, 212312, 213221, 213212, 213122, 221123, 221132, 221213, 221231, 221321, 221312, 222113, 222131, 222311, 223121, 223112, 223211, 231221, 231212, 231122, 232121, 232112, 232211, 312221, 312212, 312122, 311222, 321221, 321212, 321122, 322121, 322112, 322211]

[AABBBC, AABBCB, AABCBB, AACBBB, ABABBC, ABABCB, ABACBB, ABBABC, ABBACB, ABBBAC, ABBBCA, ABBCBA, ABBCAB, ABCBBA, ABCBAB, ABCABB, ACBBBA, ACBBAB, ACBABB, ACABBB, BAABBC, BAABCB, BAACBB, BABABC, BABACB, BABBAC, BABBCA, BABCBA, BABCAB, BACBBA, BACBAB, BACABB, BBAABC, BBAACB, BBABAC, BBABCA, BBACBA, BBACAB, BBBAAC, BBBACA, BBBCAA, BBCABA, BBCAAB, BBCBAA, BCABBA, BCABAB, BCAABB, BCBABA, BCBAAB, BCBBAA, CABBBA, CABBAB, CABABB, CAABBB, CBABBA, CBABAB, CBAABB, CBBABA, CBBAAB, CBBBAA]

zkl

<lang zkl> // eg ( (2,3,1), "ABC" ) == permute "A","A","B","B","B","C" and remove duplicates 

 //  --> ( "AABBBC", "AABBCB" .. )
 // this gets ugly lots sooner than it should

fcn permutationsWithSomeIdenticalElements(ns,abcs){ 

  ns.zipWith(fcn(n,c){ List.createLong(n,c) },abcs).flatten() : # (3,"B")-->("B","B,"B")
  Utils.Helpers.permute(_) : Utils.Helpers.listUnique(_)
  .apply("concat")  // ("A","A","B","B","B","C")-->"AABBCB"

}</lang> <lang zkl>permutationsWithSomeIdenticalElements(T(2,1),"123").println(); permutationsWithSomeIdenticalElements(T(2,1),L("\u2192","\u2191")).concat(" ").println();

z:=permutationsWithSomeIdenticalElements(T(2,3,1),"ABC"); println(z.len()); z.pump(Void,T(Void.Read,9,False), // print rows of ten items fcn{ vm.arglist.concat(" ").println() });</lang>

Output:
L("112","121","211")
→→↑  →↑→  ↑→→
60
AABBBC  AABBCB  AABCBB  AACBBB  ACABBB  CAABBB  CABABB  ACBABB  ABCABB  ABACBB
ABABCB  ABABBC  BAABBC  BAABCB  BAACBB  BACABB  BCAABB  CBAABB  ABBABC  ABBACB
ABBCAB  ABCBAB  ACBBAB  CABBAB  BABABC  BABACB  BABCAB  BACBAB  BCABAB  CBABAB
CBBAAB  BCBAAB  BBCAAB  BBACAB  BBAACB  BBAABC  CBABBA  BCABBA  BACBBA  BABCBA
BABBCA  BABBAC  CBBABA  BCBABA  BBCABA  BBACBA  BBABCA  BBABAC  ABBBAC  ABBBCA
ABBCBA  ABCBBA  ACBBBA  CABBBA  BBBAAC  BBBACA  BBBCAA  BBCBAA  BCBBAA  CBBBAA
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