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How many six-letter “words” can be formed from the alphabet $\{a – z \}$?

  • A “word” for this question must have at least one vowel $\{a, e, i, o, u\}$ and have at least one consonant (letters not in $\{a, e, i, o, u\}$ are considered as consonants).

What I have tried is $26^6−21^6$, but this does not account for the "at least one consonant case".

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    – Blue
    Jul 28, 2022 at 7:23
  • $\begingroup$ is repetition allowed ? For example , can $2a's$ and $4b's$ be used at the same time ? $\endgroup$
    – Not a Salmon Fish
    Jul 28, 2022 at 7:40
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    $\begingroup$ You subtract $21^6$ because those are all possible words with only consonants and no vowels and you want to exclude those. Do the same kind of thing for all the words with only vowels and no consonants. $\endgroup$
    – Jaap Scherphuis
    Jul 28, 2022 at 7:44
  • $\begingroup$ In addition to what @Jaap said, you need to, when properly using exclusion-inclusion, make sure that you don't doublecount. In other words, you need add the number of words that contain neither vowels nor consonants. This is because you subtracted them twice :-) $\endgroup$
    – Jyrki Lahtonen
    Jul 28, 2022 at 8:23
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    Jul 28, 2022 at 9:25

2 Answers 2

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Being a problem with two non-overlapping exclusion conditions (no absence of vowels, no absence of consonants; in a six-letter word one cannot have both absences at once) you can simply subtract both the numbers $21^6$ of words without vowels and the number $5^6$ of words without consonants from the total number of words. If there had been any overlap between the two, you would need to apply inclusion/exclusion, but that is not necessary here. By the way, might I facetiously remark that the sixth word of this sentence has $6$ distinct vowels (and $5$ distinct consonants)?

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I think that exponential generating functions can be used for this question. If there must be at least one consonant and at least one vowel , then there can be at most $5$ consonant or vowel.

  • E.G.F for consonant : $$\bigg(21x+21^2\frac{x^2}{2!}+21^3\frac{x^3}{3!}+21^4\frac{x^4}{4!}+21^5\frac{x^5}{5!}\bigg)$$

  • E.G.F for vowels : $$\bigg(5x+5^2\frac{x^2}{2!}+5^3\frac{x^3}{3!}+5^4\frac{x^4}{4!}+5^5\frac{x^5}{5!}\bigg)$$

Now find the coefficient of $\frac{x^6}{6!}$ or find $x^6$ and multiply it by $6!$ in the expansion of $$\bigg(21x+21^2\frac{x^2}{2!}+21^3\frac{x^3}{3!}+21^4\frac{x^4}{4!}+21^5\frac{x^5}{5!}\bigg)\bigg(5x+5^2\frac{x^2}{2!}+5^3\frac{x^3}{3!}+5^4\frac{x^4}{4!}+5^5\frac{x^5}{5!}\bigg)$$

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    $\begingroup$ Does one really need and exponential generating function where two subtractions will suffice? $\endgroup$
    – Marc van Leeuwen
    Jul 28, 2022 at 9:58
  • $\begingroup$ @MarcvanLeeuwen Yes , you are right , my answer is cumbersome for this question , but OP may learn different approach and apply it to more advanced question such as at least three vowels etc. $\endgroup$
    – Not a Salmon Fish
    Jul 28, 2022 at 15:42
  • $\begingroup$ For the record, the e.g.f. product has coefficients $0,0,210,8190,261870,7794150,223134030,3226532400,38717925750,398199847500,3216229537500$, the one of $x^6\over6!$ being $223134030=26^6-21^6-5^6$ indeed. The coefficient of $x^n\over n!$ is the number of words of length $n$ with this stock of letters (no more than $5$ each of vowels or consonants) and at least one vowel and one consonant. $\endgroup$
    – Marc van Leeuwen
    Jul 29, 2022 at 20:49

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