There are $21$ consonants and $5$ vowels in a language. Each word starts with a consonant and is followed by $1$ or more pairs of vowel-consonants (e.g. 'piv', 'cadit', etc.) No vowel or consonant can be repeated once it is used.
Here's how I did it:
$5! \cdot P(21, 6) = 4688409600$ different words.
I am wondering if my method is right.
Since there are 21 consonants and 5 vowels, and none of them can be used more than once, we will have up to 5 pairs after the first consonant. We have to take in account of all possibilities of pairs (1 pair, 2 pairs, 3 pairs, 4 pairs, 5 pairs).
1 Pair: 21 * 5 * 20
2 Pairs: 21 * 5 * 20 * 4 * 19
3 Pairs: 21 * 5 * 20 * 4 * 19 * 3 * 18
4 Pairs: 21 * 5 * 20 * 4 * 19 * 3 * 18 * 2 * 17
5 Pairs: 21 * 5 * 20 * 4 * 19 * 3 * 18 * 2 * 17 * 1 * 16
You would add all of these possibilities for pairs up and it would equal a final answer of 4,990,215,300