Mathematics Asked by José Pedro Ferreira on November 27, 2020

The word can have every letter as long as A,E,I,O,U aren’t all in the same word.

Can this be the number of all words of length $10$ without all vowels?

$$ 21^{10}+ sum_{i=1}^4 {10choose i} 5^i cdot 21^{10-i} + sum_{i=5}^{10} {5choose 4}{10choose i} 4^i $$

(all consonant words + all words with less than $4$ vowels + all words with $5$ nonrepeated vowels or more)

Is this correct? If there is a simpler way, I would appreciate.

The number of words in which not all the vowels appear is equal to the number of words in which at least one vowel does not appear.

Let $A, E, I, O, U$ be the set of ten-letter words in which, respectively, the letter $a, e, i, o, u$ does not appear. The set of words in which at least one vowel does not appear is $A cup E cup I cup O cup U$. By the Inclusion-Exclusion Principle, begin{align*} & |A cup E cup I cup O cup U|\ & = |A| + |E| + |I| + |O| + |U|\ & qquad - (|A cup E| + |A cup I| + |A cup O| + |A cup U| + |E cup I| + |E cup O| + |E cup U| + |I cup O| + |I cup U| + |O cup U|)\ & qquad + (|A cap E cap I| + |A cap E cap O| + |A cap E cap U| + |A cap I cap O| + |A cap I cap U| + |A cap O cap U| + |E cap I cap O| + |E cap I cap U| + |E cap O cap U| + |I cap O cap U|\ & qquad - (|A cap E cap I cap O| + |A cap E cap I cap U| + |A cap E cap O cap U| + |A cap I cap O cap U| + |E cap I cap O cap U|)\ & qquad + |A cap E cap I cap O cap U| end{align*}

$|A|$: If the vowel $a$ does not appear, then each of the ten positions can be filled in $25$ ways. Thus, $|A| = 25^{10}$.

By symmetry, $$|A| = |E| = |I| = |O| = |U|$$

$|A cap E|$: If neither of the vowels $a$ nor $e$ appears, then each of the ten positions can be filled in $24$ ways. Thus, $|A cap E| = 24^{10}$.

By symmetry, $$|A cap E| = |A cap I| = |A cap O| = |A cap U| = |E cap I| = |E cap O| = |E cap U| = |I cap O| = |I cap U| = |O cap U|$$

$|A cap E cap I|$: If none of the vowels $a, e, i$ appears, then each of the ten positions can be filled in $23$ ways. Thus, $|A cap E cap I| = 23^{10}$.

By symmetry, $$|A cap E cap I| = |A cap E cap O| = |A cap E cap U| = |A cap I cap O| = |A cap I cap U| = |A cap O cap U| = |E cap I cap O| = |E cap I cap U| = |E cap O cap U| = |I cap O cap U|$$

$|A cap E cap I cap O|$: If none of the vowels $a, e, i, o$ appears, there are $22$ ways to fill each of the ten positions. Hence, $|A cap E cap I cap O| = 22^{10}$.

By symmetry, $$|A cap E cap I cap O| = |A cap E cap I cap U| = |A cap E cap O cap U| = |A cap I cap O cap U| = |E cap I cap O cap U|$$

$|A cap E cap I cap O cap U|$: If none of the vowels $a, e, i, o, u$ appears, there are $21$ ways to fill each of the ten positions. Hence, $|A cap E cap I cap O cap U| = 21^{10}$.

Thus, by the Inclusion-Exclusion Principle, the number of ten-letter words in which at least one vowel does not appear is begin{align*} |A cup E cup I cap O cap U| & = 5 cdot 25^{10} - 10 cdot 24^{10} + 10 cdot 23^{10} - 5 cdot 22^{10} + 21^{10}\ & = binom{5}{1}25^{10} - binom{5}{2}24^{10} + binom{5}{3}23^{10} - binom{5}{4}22^{10} + binom{5}{5}21^{10}\ & = sum_{k = 1}^{5} (-1)^{k - 1}binom{5}{k}(26 - k)^{10} end{align*} where $binom{5}{k}$ is the number of ways of excluding $k$ of the $5$ vowels and $(26 - k)^{10}$ is the number of ways to form a ten-letter word with the remaining $26 - k$ letters of the alphabet.

Answered by N. F. Taussig on November 27, 2020

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