1.55 In a certain tribe's language, any sequence of 10 digits composed of 0s and 1s is a word. Two words are synonyms if and only if one word can be obtained from the other through the following procedure: first, remove several consecutive digits from the word, the sum of which should be even, and then write the removed digits in reverse order in their original positions. How many words with different meanings are there in the tribe's language?
[Solution] We refer to the digit transformation of synonyms mentioned in the problem as a synonymous transformation. If in a word, the second 1 from the right of the first 0 from the left is at least two 1s to the right, then the segment of digits from the first 0 to the second 1 from the right can be reversed, which results in the number of 1s to the right of the first 0 from the left being reduced by at least one. Therefore, each word is either 0000000000 or a word of the form
$$
\underset{n \uparrow}{11 \cdots 1} \underset{m \uparrow}{100 \cdots 0} 1 \underset{9-n \cdots \uparrow}{1}
$$
where \( n, m \geqslant 0, n+m \leqslant 9 \). It is easy to see that there are 55 words of the form (1). Therefore, there are 56 words with different meanings.

In fact, when the \( n \) values of two words of the form (1) are different, the two words are certainly not synonymous. When \( n \) is the same but \( m \) is different, because the synonymous transformation does not change the value of \( m \), the two words are also not synonymous.