Create all possible words using a set or lettersFinding 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
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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
$begingroup$
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
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
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
edited 1 hour ago
J. M. is slightly pensive♦
98.3k10306466
asked 1 hour ago
mf67mf67
975
$endgroup$
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
string-manipulation combinatorics
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
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
edited 1 hour ago
J. M. is slightly pensive♦
98.3k10306466
98.3k10306466
asked 1 hour ago
mf67mf67
975
asked 1 hour ago
mf67mf67
975
asked 1 hour ago
mf67mf67
975
975
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
answered 1 hour ago
LeeLee
46027
$endgroup$
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"
answered 1 hour ago
bill sbill s
54.6k377156
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered 1 hour ago
LeeLee
46027
$endgroup$
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.
answered 1 hour ago
LeeLee
46027
$endgroup$
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.
answered 1 hour ago
LeeLee
46027
$endgroup$
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.
answered 1 hour ago
LeeLee
46027
answered 1 hour ago
LeeLee
46027
answered 1 hour ago
LeeLee
46027
answered 1 hour ago
LeeLee
46027
46027
add a comment |
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"
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"
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"
answered 1 hour ago
bill sbill s
54.6k377156
$endgroup$
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"
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
answered 1 hour ago
bill sbill s
54.6k377156
54.6k377156
add a comment |
add a comment |
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