How many six-letter "words" can be formed from the alphabet $\{a-z\}$, if a "word" contains at least one vowel and at least one consonant? 
How many six-letter “words” can be formed from the alphabet $\{a – z \}$?

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*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".
 A: 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)?
A: 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.

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*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 $$\bigg(21x+21^2\frac{x^2}{2!}+21^3\frac{x^3}{3!}+21^4\frac{x^4}{4!}+21^5\frac{x^5}{5!}\bigg)\bigg(5x+5^2\frac{x^2}{2!}+5^3\frac{x^3}{3!}+5^4\frac{x^4}{4!}+5^5\frac{x^5}{5!}\bigg)$$
