How many possible word combinations can exist for a alphabet of 4 vowels and 8 consonants, 12 letters altogether where a word can start with a vowel or consonant, but every word must end in a vowel? Moreover, every vowel must be followed by a consonant (except at the end). Each consonant can be followed by another consonant, or a vowel. Double consonants must be followed by a vowel. Max letters per word are six. How many possible words can exist?
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2$\begingroup$ What have you tried ? It is small enough to enumerate patterns of consonants and vowels $\endgroup$– true blue anilNov 15, 2017 at 4:19
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$\begingroup$ The rules that every word must end in a vowel and that every vowel must be followed by a consonant contradict each other. $\endgroup$– N. F. TaussigNov 15, 2017 at 9:24
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$\begingroup$ @N.F.Taussig I edited it and added "except at the end". $\endgroup$– XarcellNov 15, 2017 at 19:12
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$\begingroup$ @trueblueanil I'm not very good with math. I can come up with an estimation in my head, but don't know to write it down. I came up with 380 words, which doesn't sound anywhere close. $\endgroup$– XarcellNov 15, 2017 at 19:14
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2 Answers
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First write down strings like VCCVCV that represent allowable words where V is a vowel and C is a consonant. Without the restrictions there would be $2^6=64$ of them. Ending in a vowel cuts that in half, so there are less than $32$ of them. For each one use the multiplication principle. Each V gives $4$ possibilities, each C gives $8$, so VCCVCV gives $4^38^3=2^{15}=32768$ words. You just need to find how many patters there are with $3$ vowels and $3$ consonants, then mulitply by $32768$ and add to the results with other numbers of vowels/consonants.
answered Nov 15, 2017 at 19:19
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Find, by chasing cases, the allowed words of length $6$ over the alphabet $\{V,C\}$. Sort them into types according to the number of vowels occurring. Then for each type determine in how many ways you can distribute the available $4$ vowels and $8$ consonants on a $\{V,C\}$-word of this type.
A word can begin with $VC$, $CVC$, or $CCVC$, and has to end with $CV$. This leaves the four types $$VCCVCV,\quad VCVCCV,\quad CVCVCV,\quad CCVCCV\ ,$$ three of them of type $3V\ 3C$, and one of them of type $2V\ 4C$. It follows that there are $$3\cdot4^3\cdot8^3+ 4^2\cdot8^4=163\,840$$ admissible words.
answered Nov 15, 2017 at 19:20