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Create all possible words using a set or letters

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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 3 hours ago

J. M. is slightly pensive♦

98.3k10307466

asked 3 hours ago

mf67mf67

976

$endgroup$

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 3 hours ago

J. M. is slightly pensive♦

98.3k10307466

asked 3 hours ago

mf67mf67

976

$endgroup$

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 3 hours ago

J. M. is slightly pensive♦

98.3k10307466

asked 3 hours ago

mf67mf67

976

$endgroup$

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

share|improve this question

edited 3 hours ago

J. M. is slightly pensive♦

98.3k10307466

asked 3 hours ago

mf67mf67

976

share|improve this question

edited 3 hours ago

J. M. is slightly pensive♦

98.3k10307466

asked 3 hours ago

mf67mf67

976

edited 3 hours ago

J. M. is slightly pensive♦

98.3k10307466

edited 3 hours ago

J. M. is slightly pensive♦

98.3k10307466

asked 3 hours ago

mf67mf67

976

asked 3 hours ago

mf67mf67

976

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 3 hours 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
  1 hour 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 3 hours ago

bill sbill s

54.6k377157

$endgroup$

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|improve this answer

answered 1 hour 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 3 hours 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
  1 hour 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 3 hours 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
  1 hour 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 3 hours ago

LeeLee

46027

$endgroup$

StringJoin /@ Permutations[letters]

Permutations[letters]

share|improve this answer

answered 3 hours ago

LeeLee

46027

answered 3 hours ago

LeeLee

46027

answered 3 hours 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 3 hours ago

bill sbill s

54.6k377157

$endgroup$

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 3 hours ago

bill sbill s

54.6k377157

$endgroup$

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 3 hours ago

bill sbill s

54.6k377157

$endgroup$

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 3 hours ago

bill sbill s

54.6k377157

answered 3 hours ago

bill sbill s

54.6k377157

answered 3 hours ago

bill sbill s

54.6k377157

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|improve this answer

answered 1 hour 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|improve this answer

answered 1 hour 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|improve this answer

answered 1 hour 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|improve this answer

answered 1 hour ago

JagraJagra

7,85312159

answered 1 hour ago

JagraJagra

7,85312159

answered 1 hour ago

JagraJagra

7,85312159