The estimate $100 000^{10}-1$ comes from assuming a discrete model for the $10$ consecutive words, without any simplifications or restrictions, thus using all interactions up to and including order $10$.
It is not important that the words are consecutive, we would get the same count for any ten specified word positions. For each position, it can be any of the $100000$ words, so we need that number of probabilities. So you can build up a cube in $10$-space, wity each dimension cut up in $100000$ boxes. Taking all the combinations, that give $100000^{10}$ boxes, each box giving on possible $10$-word sequence, such as " am I writing now holy blue crap green integrated ideas", which would be a sequence of fairly low probability. Then subtract $1$ to account for the fact that the probabilities must sum to one!
In practice, a fairly high number of the probabilities would be zero, because they correspond to non-grammatical utterings.