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The english alphabet contains $21$ consonants and $5$ vowels. How many strings of $6$ lowercase letters of the English alphabet contain

  • a) exactly 1 vowel
  • b) exactly 2 vowels
  • c) at least one vowel
  • d) at least two vowels
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What are your thoughts on how to approach it? Where are you getting stuck? If this is homework, you should tag it as such. –  Jonathan Christensen Nov 28 '12 at 18:51

1 Answer 1

Exactly $1$ vowel: The location of the vowel can be chosen in $\binom{6}{1}$ ways. (Of course this is $6$, but we are trying to use a technique that works more generally.)

For each choice of location, the location can be filled with a vowel in $5$ ways.

That leaves $5$ empty locations, which can be filled with consonants in $21^5$ ways. That gives a total of $$\binom{6}{1}(5)(21^5).$$

Exactly $2$ vowels: The location of the vowels can be chosen in $\binom{6}{2}$ ways.

For each choice of locations, the locations can be filled with vowels in $5^2$ ways.

That leaves $4$ empty locations, which can be filled with consonants in $21^4$ ways. That gives a total of $$\binom{6}{2}(5^2)(21^4).$$

At least $1$ vowel: There are $26^6$ $6$-letter "words." And there are $21^6$ all consonant words. So there are $26^6-21^6$ words that have at least one vowel.

There are other ways of counting this, but they are less efficient.

At least $2$ vowels: Again, there are various ways of counting. An efficient way is to count the words that have $0$ vowels or $1$ vowels, and subtract from the total number of words. This approach lets us recycle previous results, and recycling is a virtue.

So take the total number $26^6$ of words, and subtract the $21^6$ all consonant words, and the number of $1$-vowel words we already calculated.

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