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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 1 hour ago

J. M. is slightly pensive♦

98.3k10306466

asked 1 hour 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 1 hour 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 1 hour ago

mf67mf67

975

$endgroup$

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

share|improve this question

edited 1 hour ago

J. M. is slightly pensive♦

98.3k10306466

asked 1 hour ago

mf67mf67

975

share|improve this question

edited 1 hour ago

J. M. is slightly pensive♦

98.3k10306466

asked 1 hour 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 1 hour ago

mf67mf67

975

asked 1 hour ago

mf67mf67

975

2 Answers
2

active

oldest

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2

$begingroup$

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

45027

$endgroup$

add a comment | 

1

$begingroup$

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 | 

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2

$begingroup$

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

45027

$endgroup$

add a comment | 

2

$begingroup$

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

45027

$endgroup$

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

45027

$endgroup$

StringJoin /@ Permutations[letters]

Permutations[letters]

share|improve this answer

answered 1 hour ago

LeeLee

45027

answered 1 hour ago

LeeLee

45027

answered 1 hour ago

LeeLee

45027

1

$begingroup$

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 | 

1

$begingroup$

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