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4

$begingroup$

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 2 days ago

J. M. is slightly pensive♦

98.5k10308466

asked 2 days ago

mf67mf67

1126

$endgroup$

add a comment | 

4

$begingroup$

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 2 days ago

J. M. is slightly pensive♦

98.5k10308466

asked 2 days ago

mf67mf67

1126

$endgroup$

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 2 days ago

J. M. is slightly pensive♦

98.5k10308466

asked 2 days ago

mf67mf67

1126

$endgroup$

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

share|improve this question

edited 2 days ago

J. M. is slightly pensive♦

98.5k10308466

asked 2 days ago

mf67mf67

1126

share|improve this question

edited 2 days ago

J. M. is slightly pensive♦

98.5k10308466

asked 2 days ago

mf67mf67

1126

edited 2 days ago

J. M. is slightly pensive♦

98.5k10308466

edited 2 days ago

J. M. is slightly pensive♦

98.5k10308466

asked 2 days ago

mf67mf67

1126

asked 2 days ago

mf67mf67

1126

3 Answers
3

active

oldest

votes

9

$begingroup$

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"}}

To get all six-letter words:

letters = {"a", "b", "c", "d", "e", "f"};

perms = Permutations[letters, {6}];

StringJoin /@ perms



{"abcdef", "abcdfe", "abcedf", "abcefd", "abcfde" ... etc.

there are a lot of them.

share|improve this answer

edited yesterday

answered 2 days ago

bill sbill s

54.7k377157

$endgroup$

add a comment | 

7

$begingroup$

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 2 days ago

LeeLee

50027

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  yesterday
  

  
  
  
  

add a comment | 

3

$begingroup$

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 yesterday

JagraJagra

7,88312159

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  No, the OP requested all six letter words.
  $endgroup$
  – m_goldberg
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Your last line can be written more cleanly as Map@StringJoin/@p or Map[StringJoin, p, {2}].
  $endgroup$
  – Doorknob
  yesterday
  

  
  
  
  

add a comment | 

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9

$begingroup$

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"}}

To get all six-letter words:

letters = {"a", "b", "c", "d", "e", "f"};

perms = Permutations[letters, {6}];

StringJoin /@ perms



{"abcdef", "abcdfe", "abcedf", "abcefd", "abcfde" ... etc.

there are a lot of them.

share|improve this answer

edited yesterday

answered 2 days ago

bill sbill s

54.7k377157

$endgroup$

add a comment | 

9

$begingroup$

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"}}

To get all six-letter words:

letters = {"a", "b", "c", "d", "e", "f"};

perms = Permutations[letters, {6}];

StringJoin /@ perms



{"abcdef", "abcdfe", "abcedf", "abcefd", "abcfde" ... etc.

there are a lot of them.

share|improve this answer

edited yesterday

answered 2 days ago

bill sbill s

54.7k377157

$endgroup$

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"}}

To get all six-letter words:

letters = {"a", "b", "c", "d", "e", "f"};

perms = Permutations[letters, {6}];

StringJoin /@ perms



{"abcdef", "abcdfe", "abcedf", "abcefd", "abcfde" ... etc.

there are a lot of them.

share|improve this answer

edited yesterday

answered 2 days ago

bill sbill s

54.7k377157

$endgroup$

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"}}

letters = {"a", "b", "c", "d", "e", "f"};

perms = Permutations[letters, {6}];

StringJoin /@ perms



{"abcdef", "abcdfe", "abcedf", "abcefd", "abcfde" ... etc.

share|improve this answer

edited yesterday

answered 2 days ago

bill sbill s

54.7k377157

answered 2 days ago

bill sbill s

54.7k377157

answered 2 days ago

bill sbill s

54.7k377157

7

$begingroup$

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 2 days ago

LeeLee

50027

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  yesterday
  

  
  
  
  

add a comment | 

7

$begingroup$

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 2 days ago

LeeLee

50027

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  yesterday
  

  
  
  
  

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 2 days ago

LeeLee

50027

$endgroup$

StringJoin /@ Permutations[letters]

Permutations[letters]

share|improve this answer

answered 2 days ago

LeeLee

50027

answered 2 days ago

LeeLee

50027

answered 2 days ago

LeeLee

50027

3

$begingroup$

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 yesterday

JagraJagra

7,88312159

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  No, the OP requested all six letter words.
  $endgroup$
  – m_goldberg
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Your last line can be written more cleanly as Map@StringJoin/@p or Map[StringJoin, p, {2}].
  $endgroup$
  – Doorknob
  yesterday
  

  
  
  
  

add a comment | 

3

$begingroup$

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 yesterday

JagraJagra

7,88312159

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  No, the OP requested all six letter words.
  $endgroup$
  – m_goldberg
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Your last line can be written more cleanly as Map@StringJoin/@p or Map[StringJoin, p, {2}].
  $endgroup$
  – Doorknob
  yesterday
  

  
  
  
  

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 yesterday

JagraJagra

7,88312159

$endgroup$

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 yesterday

JagraJagra

7,88312159

answered yesterday

JagraJagra

7,88312159

answered yesterday

JagraJagra

7,88312159