1. In a certain country, the alphabet consists of three letters: "M", "G", and "U". A word is any finite sequence of these letters, in which two consonants cannot stand next to each other and two vowels cannot stand next to each other. How many 200-letter words are there in this country that contain each of the three letters at least once?
Answer: $2^{101}-4$. Solution. After understanding the structure of the words, we consider three cases.

1) The first letter is M. Then the second is U, the third is G or M, the fourth is U, the fifth is G or M, ..., the 200th is U. There will be $2^{99}$ such words. 2) Similarly, if the first letter is G.
2) The first letter is U. Then the second is G or M, the third is U, the fourth is G or M, ..., the 200th is G or M. There will be $2^{100}$ such words.

In total, there are $2^{99}+2^{99}+2^{100}=2^{101}$ words. From this, we need to subtract 4 words: MUMU....MUM, GUGU...GUG, UMUM...UMU, and UGUG...UGU (each of these words lacks one of the letters).