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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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**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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