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Create all possible words using a set or letters

Finding all length-n words on an alphabet that have a specified number of each letterFinding all dictionary words that can be made with a given set of characters (Wordfeud/Scrabble)How to enumerate all possible binary associations?Sorting an Array with words in different languagesUsing StringCases and treating certain phrases as single wordsGraph showing valid English words obtained by insertion of single charactersTrim a list of elementsList all possible microstates and corresponding energy using mathematica.Selecting words having a specific number of letters from a textSelecting elements using two lists

1

$begingroup$

Given a list of letters,

letters = "A", "B", ..., "F" 

is it possible to get Mathematica to generate all ‘words’ (in this example, 6 letter words), if only one letter can be used one time only, e.g. ABCDEF, ABCDFE, …? TIA.

string-manipulation combinatorics

share|improve this question

edited 1 hour ago

J. M. is slightly pensive♦

98.3k10306466

asked 2 hours ago

mf67mf67

975

$endgroup$

add a comment | 

1

$begingroup$

Given a list of letters,

letters = "A", "B", ..., "F" 

is it possible to get Mathematica to generate all ‘words’ (in this example, 6 letter words), if only one letter can be used one time only, e.g. ABCDEF, ABCDFE, …? TIA.

string-manipulation combinatorics

share|improve this question

edited 1 hour ago

J. M. is slightly pensive♦

98.3k10306466

asked 2 hours ago

mf67mf67

975

$endgroup$

add a comment | 

$begingroup$

Given a list of letters,

letters = "A", "B", ..., "F" 

is it possible to get Mathematica to generate all ‘words’ (in this example, 6 letter words), if only one letter can be used one time only, e.g. ABCDEF, ABCDFE, …? TIA.

string-manipulation combinatorics

share|improve this question

edited 1 hour ago

J. M. is slightly pensive♦

98.3k10306466

asked 2 hours ago

mf67mf67

975

$endgroup$

letters = "A", "B", ..., "F" 

share|improve this question

edited 1 hour ago

J. M. is slightly pensive♦

98.3k10306466

asked 2 hours ago

mf67mf67

975

share|improve this question

edited 1 hour ago

J. M. is slightly pensive♦

98.3k10306466

asked 2 hours ago

mf67mf67

975

edited 1 hour ago

J. M. is slightly pensive♦

98.3k10306466

edited 1 hour ago

J. M. is slightly pensive♦

98.3k10306466

asked 2 hours ago

mf67mf67

975

asked 2 hours ago

mf67mf67

975

3 Answers
3

active

oldest

votes

3

$begingroup$

You can create permutations with all of the letters as strings with:

StringJoin /@ Permutations[letters]

If you want lists of the individual letters just use:

Permutations[letters]

Check the documentation of Permutations to learn about permutations with subsets of letters. If you want to use each letter more than once, look at the documentation for Tuples.

share|improve this answer

answered 1 hour ago

LeeLee

46027

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks(x2). Is there some way to check how many words contain a ‘sub-word’, like ‘ab’ or even a set of ‘sub-words’ like ‘ab’ and ‘cd’? And is there any web page or text book that deals with combinatorics in Mathematica (on a more ‘basic’ level) that I could visit/buy and read?
  $endgroup$
  – mf67
  2 mins ago
  

  
  
  
  

add a comment | 

2

$begingroup$

Pemutations will do it:

letters = "a", "b", "c";
Permutations[letters, 3]
"a", "b", "c", "a", "c", "b", "b", "a", "c", 
 "b", "c", "a", "c", "a", "b", "c", "b", "a"

share|improve this answer

answered 1 hour ago

bill sbill s

54.6k377156

$endgroup$

add a comment | 

0

$begingroup$

If I follow the OP's question, I think they want the following:

letters = "a", "b", "c";
p = Permutations[letters, #] & /@ Range[Length[letters]];
(StringJoin[#] & /@ #) & /@ p

a, b, c, ab, ac, ba, bc, ca, cb, abc, acb, bac, bca, cab, cba

share

answered 5 mins ago

JagraJagra

7,85312159

$endgroup$

add a comment | 

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3

$begingroup$

You can create permutations with all of the letters as strings with:

StringJoin /@ Permutations[letters]

If you want lists of the individual letters just use:

Permutations[letters]

Check the documentation of Permutations to learn about permutations with subsets of letters. If you want to use each letter more than once, look at the documentation for Tuples.

share|improve this answer

answered 1 hour ago

LeeLee

46027

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks(x2). Is there some way to check how many words contain a ‘sub-word’, like ‘ab’ or even a set of ‘sub-words’ like ‘ab’ and ‘cd’? And is there any web page or text book that deals with combinatorics in Mathematica (on a more ‘basic’ level) that I could visit/buy and read?
  $endgroup$
  – mf67
  2 mins ago
  

  
  
  
  

add a comment | 

3

$begingroup$

You can create permutations with all of the letters as strings with:

StringJoin /@ Permutations[letters]

If you want lists of the individual letters just use:

Permutations[letters]

Check the documentation of Permutations to learn about permutations with subsets of letters. If you want to use each letter more than once, look at the documentation for Tuples.

share|improve this answer

answered 1 hour ago

LeeLee

46027

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks(x2). Is there some way to check how many words contain a ‘sub-word’, like ‘ab’ or even a set of ‘sub-words’ like ‘ab’ and ‘cd’? And is there any web page or text book that deals with combinatorics in Mathematica (on a more ‘basic’ level) that I could visit/buy and read?
  $endgroup$
  – mf67
  2 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

You can create permutations with all of the letters as strings with:

StringJoin /@ Permutations[letters]

If you want lists of the individual letters just use:

Permutations[letters]

Check the documentation of Permutations to learn about permutations with subsets of letters. If you want to use each letter more than once, look at the documentation for Tuples.

share|improve this answer

answered 1 hour ago

LeeLee

46027

$endgroup$

StringJoin /@ Permutations[letters]

Permutations[letters]

share|improve this answer

answered 1 hour ago

LeeLee

46027

answered 1 hour ago

LeeLee

46027

answered 1 hour ago

LeeLee

46027

2

$begingroup$

Pemutations will do it:

letters = "a", "b", "c";
Permutations[letters, 3]
"a", "b", "c", "a", "c", "b", "b", "a", "c", 
 "b", "c", "a", "c", "a", "b", "c", "b", "a"

share|improve this answer

answered 1 hour ago

bill sbill s

54.6k377156

$endgroup$

add a comment | 

2

$begingroup$

Pemutations will do it:

letters = "a", "b", "c";
Permutations[letters, 3]
"a", "b", "c", "a", "c", "b", "b", "a", "c", 
 "b", "c", "a", "c", "a", "b", "c", "b", "a"

share|improve this answer

answered 1 hour ago

bill sbill s

54.6k377156

$endgroup$

add a comment | 

$begingroup$

Pemutations will do it:

letters = "a", "b", "c";
Permutations[letters, 3]
"a", "b", "c", "a", "c", "b", "b", "a", "c", 
 "b", "c", "a", "c", "a", "b", "c", "b", "a"

share|improve this answer

answered 1 hour ago

bill sbill s

54.6k377156

$endgroup$

letters = "a", "b", "c";
Permutations[letters, 3]
"a", "b", "c", "a", "c", "b", "b", "a", "c", 
 "b", "c", "a", "c", "a", "b", "c", "b", "a"

share|improve this answer

answered 1 hour ago

bill sbill s

54.6k377156

answered 1 hour ago

bill sbill s

54.6k377156

answered 1 hour ago

bill sbill s

54.6k377156

0

$begingroup$

If I follow the OP's question, I think they want the following:

letters = "a", "b", "c";
p = Permutations[letters, #] & /@ Range[Length[letters]];
(StringJoin[#] & /@ #) & /@ p

a, b, c, ab, ac, ba, bc, ca, cb, abc, acb, bac, bca, cab, cba

share

answered 5 mins ago

JagraJagra

7,85312159

$endgroup$

add a comment | 

0

$begingroup$

If I follow the OP's question, I think they want the following:

letters = "a", "b", "c";
p = Permutations[letters, #] & /@ Range[Length[letters]];
(StringJoin[#] & /@ #) & /@ p

a, b, c, ab, ac, ba, bc, ca, cb, abc, acb, bac, bca, cab, cba

share

answered 5 mins ago

JagraJagra

7,85312159

$endgroup$

add a comment | 

$begingroup$

If I follow the OP's question, I think they want the following:

letters = "a", "b", "c";
p = Permutations[letters, #] & /@ Range[Length[letters]];
(StringJoin[#] & /@ #) & /@ p

a, b, c, ab, ac, ba, bc, ca, cb, abc, acb, bac, bca, cab, cba

share

answered 5 mins ago

JagraJagra

7,85312159

$endgroup$

letters = "a", "b", "c";
p = Permutations[letters, #] & /@ Range[Length[letters]];
(StringJoin[#] & /@ #) & /@ p

a, b, c, ab, ac, ba, bc, ca, cb, abc, acb, bac, bca, cab, cba

share

answered 5 mins ago

JagraJagra

7,85312159

answered 5 mins ago

JagraJagra

7,85312159

answered 5 mins ago

JagraJagra

7,85312159