A new language uses only the letters A, B, C, D, and E. The letters A and E are called vowels, while the letters B, C and D are called consonants. A sequence of letters is called a word if it does not include the same letter twice in a row, and it does not include two vowels in a row. How many words are there in this language that are 10 letters long and that begin with a vowel?
(A) 199680
(B) 199968
(C) 199584
(D) 199872
(E) 199776
Using the given rules, the words that are 1 letter long are A, B, C, D, E.

Using the given rules, the words that are 2 letters long are $\mathrm{AB}, \mathrm{AC}, \mathrm{AD}, \mathrm{BA}, \mathrm{BC}, \mathrm{BD}, \mathrm{BE}$, $\mathrm{CA}, \mathrm{CB}, \mathrm{CD}, \mathrm{CE}, \mathrm{DA}, \mathrm{DB}, \mathrm{DC}, \mathrm{DE}, \mathrm{EB}, \mathrm{EC}, \mathrm{ED}$.

Let $v_{n}$ be the number of words that are $n$ letters long and that begin with a vowel. Note that $v_{1}=2$ and $v_{2}=6$.

Let $c_{n}$ be the number of words that are $n$ letters long and that begin with a consonant. Note that $c_{1}=3$ and $c_{2}=12$.

Suppose $n \geq 2$.

Consider a word of length $n$ that begins with a vowel (that is, with A or E).

Since two vowels cannot occur in a row, the second letter of this word must be B, C or D.

This means that every word of length $n$ that begins with a vowel can be transformed into a word of length $n-1$ that begins with a consonant by removing the first letter.

Also, each word of length $n-1$ that begins with a consonant can form two different words of length $n$ that begin with a vowel.

Therefore, $v_{n}=2 c_{n-1}$.

Consider a word of length $n$ that begins with a consonant.

Since the same letter cannot occur twice in a row, then the second letter of this word is either a vowel or a different consonant than the first letter of the word.

Each word of length $n-1$ that begins with a vowel can form 3 words of length $n$ that begin with a consonant, obtained by adding B, C, D to the start of the word.

Each word of length $n-1$ that begins with a consonant can form 2 words of length $n$ that begin with a consonant, obtained by adding each of the consonants other than the one with which the word of length $n-1$ starts.

Therefore, $c_{n}=3 v_{n-1}+2 c_{n-1}$.

We note from above that $v_{1}=2$ and $c_{1}=3$.

The equations $v_{2}=2 c_{1}$ and $c_{2}=3 v_{1}+2 c_{1}$ are consistent with the information that $v_{2}=6$ and $c_{2}=12$.

Since $v_{2}=6$ and $c_{2}=12$, then $v_{3}=2 c_{2}=24$ and $c_{3}=3 v_{2}+2 c_{2}=3(6)+2(12)=42$.

We want to determine $v_{10}$.

We continue these calculations and make a table:

| $n$ | $v_{n}$ | $c_{n}$ |
| :---: | :---: | :---: |
| 1 | 2 | 3 |
| 2 | 6 | 12 |
| 3 | 24 | 42 |
| 4 | 84 | 156 |
| 5 | 312 | 564 |
| 6 | 1128 | 2064 |
| 7 | 4128 | 7512 |
| 8 | 15024 | 27408 |
| 9 | 54816 | 99888 |
| 10 | 199776 | 364224 |

Therefore, there are 199776 words of length 10 that begin with a vowel.

ANSWER: (E)