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Counting the number of strings starting with “t” and containing 2 vowels and 2 consonants

Counting the number of a specific type of permutationHow can I generate permutations of bit strings with repetition?Counting number of inversionsCounting the total number of instances of some collection of objects in a listDetermine repeating number of a given number and corresponding positionDetermine the number of times a function will be calledCounting number of spheres that contain a pointCounting the permuted partitionsCounting the number of Configurations in an Array with ConstraintsCount sublists containing certain strings

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0;
}

3

$begingroup$

I generated a list of strings that match the criteria in the title as follows:

vowels = Characters @ "aeiou";

consonants = DeleteCases[Complement[Alphabet[], vowels], "t"];



v = Select[Tuples[vowels, {2}], Length @ Tally @ # != 1];

c = Select[Tuples[consonants, {2}], Length @ Tally @ # != 1];



Flatten[#, 2] & @

    Table[

        Permutations[Flatten[{v[[i]], c[[j]]}], {4}],

        {i, Length @ v}, {j, Length @ c}

    ];



words = DeleteDuplicates["t" <> # & /@ %];

Length @ words

(* 45600 *)

The count agrees with my math; there are $binom52$ ways of picking vowels, $binom{20}2$ ways of picking (non-t) consonants, and $4!$ ways of arranging all but the first letter, hence $4!binom52binom{20}2=45600$ total words.

Is there a better way of generating all possible strings with the given criteria? I don't see a problem with my method, but think there is some corner-cutting that can be done.

combinatorics permutation counting

share|improve this question

asked 8 hours ago

user170231user170231

73755 silver badges1313 bronze badges

$endgroup$

add a comment | 

3

$begingroup$

I generated a list of strings that match the criteria in the title as follows:

vowels = Characters @ "aeiou";

consonants = DeleteCases[Complement[Alphabet[], vowels], "t"];



v = Select[Tuples[vowels, {2}], Length @ Tally @ # != 1];

c = Select[Tuples[consonants, {2}], Length @ Tally @ # != 1];



Flatten[#, 2] & @

    Table[

        Permutations[Flatten[{v[[i]], c[[j]]}], {4}],

        {i, Length @ v}, {j, Length @ c}

    ];



words = DeleteDuplicates["t" <> # & /@ %];

Length @ words

(* 45600 *)

The count agrees with my math; there are $binom52$ ways of picking vowels, $binom{20}2$ ways of picking (non-t) consonants, and $4!$ ways of arranging all but the first letter, hence $4!binom52binom{20}2=45600$ total words.

Is there a better way of generating all possible strings with the given criteria? I don't see a problem with my method, but think there is some corner-cutting that can be done.

combinatorics permutation counting

share|improve this question

asked 8 hours ago

user170231user170231

73755 silver badges1313 bronze badges

$endgroup$

add a comment | 

$begingroup$

I generated a list of strings that match the criteria in the title as follows:

vowels = Characters @ "aeiou";

consonants = DeleteCases[Complement[Alphabet[], vowels], "t"];



v = Select[Tuples[vowels, {2}], Length @ Tally @ # != 1];

c = Select[Tuples[consonants, {2}], Length @ Tally @ # != 1];



Flatten[#, 2] & @

    Table[

        Permutations[Flatten[{v[[i]], c[[j]]}], {4}],

        {i, Length @ v}, {j, Length @ c}

    ];



words = DeleteDuplicates["t" <> # & /@ %];

Length @ words

(* 45600 *)

The count agrees with my math; there are $binom52$ ways of picking vowels, $binom{20}2$ ways of picking (non-t) consonants, and $4!$ ways of arranging all but the first letter, hence $4!binom52binom{20}2=45600$ total words.

Is there a better way of generating all possible strings with the given criteria? I don't see a problem with my method, but think there is some corner-cutting that can be done.

combinatorics permutation counting

share|improve this question

asked 8 hours ago

user170231user170231

73755 silver badges1313 bronze badges

$endgroup$

vowels = Characters @ "aeiou";

consonants = DeleteCases[Complement[Alphabet[], vowels], "t"];



v = Select[Tuples[vowels, {2}], Length @ Tally @ # != 1];

c = Select[Tuples[consonants, {2}], Length @ Tally @ # != 1];



Flatten[#, 2] & @

    Table[

        Permutations[Flatten[{v[[i]], c[[j]]}], {4}],

        {i, Length @ v}, {j, Length @ c}

    ];



words = DeleteDuplicates["t" <> # & /@ %];

Length @ words

(* 45600 *)

share|improve this question

asked 8 hours ago

user170231user170231

73755 silver badges1313 bronze badges

share|improve this question

asked 8 hours ago

user170231user170231

73755 silver badges1313 bronze badges

asked 8 hours ago

user170231user170231

73755 silver badges1313 bronze badges

asked 8 hours ago

user170231user170231

73755 silver badges1313 bronze badges

1 Answer
1

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4

$begingroup$

For "counting the number of strings ..." you don't have to generate the list of such strings:

4! Binomial[Length @ vowels, 2] Binomial[Length @ consonants, 2]

45600

An alternative way to generate the list of words:

words = StringJoin["t", ##] & @@@ (Join @@ (Permutations /@ (Join @@@ 

         Tuples[Subsets[#, {2}] & /@ {vowels, consonants}])));



Length @ words

45600

words // Short

{taebc, taecb, tabec, tabce, taceb, << 45591 >>, tzuoy, tzuyo, tzyou, tzyuo}

share|improve this answer

edited 8 hours ago

answered 8 hours ago

kglrkglr

214k1010 gold badges244244 silver badges488488 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Right, I was just trying to make an exercise in WL to confirm this count :)
  $endgroup$
  – user170231
  8 hours ago
  

  
  
  
  

add a comment | 

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4

$begingroup$

For "counting the number of strings ..." you don't have to generate the list of such strings:

4! Binomial[Length @ vowels, 2] Binomial[Length @ consonants, 2]

45600

An alternative way to generate the list of words:

words = StringJoin["t", ##] & @@@ (Join @@ (Permutations /@ (Join @@@ 

         Tuples[Subsets[#, {2}] & /@ {vowels, consonants}])));



Length @ words

45600

words // Short

{taebc, taecb, tabec, tabce, taceb, << 45591 >>, tzuoy, tzuyo, tzyou, tzyuo}

share|improve this answer

edited 8 hours ago

answered 8 hours ago

kglrkglr

214k1010 gold badges244244 silver badges488488 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Right, I was just trying to make an exercise in WL to confirm this count :)
  $endgroup$
  – user170231
  8 hours ago
  

  
  
  
  

add a comment | 

4

$begingroup$

For "counting the number of strings ..." you don't have to generate the list of such strings:

4! Binomial[Length @ vowels, 2] Binomial[Length @ consonants, 2]

45600

An alternative way to generate the list of words:

words = StringJoin["t", ##] & @@@ (Join @@ (Permutations /@ (Join @@@ 

         Tuples[Subsets[#, {2}] & /@ {vowels, consonants}])));



Length @ words

45600

words // Short

{taebc, taecb, tabec, tabce, taceb, << 45591 >>, tzuoy, tzuyo, tzyou, tzyuo}

share|improve this answer

edited 8 hours ago

answered 8 hours ago

kglrkglr

214k1010 gold badges244244 silver badges488488 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Right, I was just trying to make an exercise in WL to confirm this count :)
  $endgroup$
  – user170231
  8 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

For "counting the number of strings ..." you don't have to generate the list of such strings:

4! Binomial[Length @ vowels, 2] Binomial[Length @ consonants, 2]

45600

An alternative way to generate the list of words:

words = StringJoin["t", ##] & @@@ (Join @@ (Permutations /@ (Join @@@ 

         Tuples[Subsets[#, {2}] & /@ {vowels, consonants}])));



Length @ words

45600

words // Short

{taebc, taecb, tabec, tabce, taceb, << 45591 >>, tzuoy, tzuyo, tzyou, tzyuo}

share|improve this answer

edited 8 hours ago

answered 8 hours ago

kglrkglr

214k1010 gold badges244244 silver badges488488 bronze badges

$endgroup$

4! Binomial[Length @ vowels, 2] Binomial[Length @ consonants, 2]

words = StringJoin["t", ##] & @@@ (Join @@ (Permutations /@ (Join @@@ 

         Tuples[Subsets[#, {2}] & /@ {vowels, consonants}])));



Length @ words

words // Short

share|improve this answer

edited 8 hours ago

answered 8 hours ago

kglrkglr

214k1010 gold badges244244 silver badges488488 bronze badges

answered 8 hours ago

kglrkglr

214k1010 gold badges244244 silver badges488488 bronze badges

answered 8 hours ago

kglrkglr

214k1010 gold badges244244 silver badges488488 bronze badges