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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
$endgroup$
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
$endgroup$
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
$endgroup$
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
98.3k10306466
asked 2 hours ago
mf67mf67
975
975
add a comment |
add a comment |
3 Answers
3
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.
$endgroup$
$begingroup$
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
19 mins ago
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"
$endgroup$
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
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
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.
$endgroup$
$begingroup$
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
19 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.
$endgroup$
$begingroup$
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
19 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.
$endgroup$
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
46027
$begingroup$
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
19 mins ago
add a comment |
$begingroup$
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
19 mins ago
$begingroup$
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
19 mins ago
$begingroup$
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
19 mins ago
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"
$endgroup$
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"
$endgroup$
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"
$endgroup$
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 2 hours ago
bill sbill s
54.6k377156
54.6k377156
add a comment |
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
$endgroup$
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
$endgroup$
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
$endgroup$
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
answered 22 mins ago
JagraJagra
7,85312159
7,85312159
add a comment |
add a comment |
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