you can do the maths. if the words are drawn from a dictionary of size D and you use N words then the number of choices that need to be explored is D^N. the equivalent number of bits is log2 of that.

so what's D? i imagine most people would use common words, so let's take the most popular 1 percent of "dict/words" on my computer. that's 4,000 words. log2(D^N) = N * log2(D) = N * 12 (since log2(1000)~10).

now in another thread here there's an article saying that 9 characters drawn from an alphabet of 100 (lower+upper+symbols) is around the current limit. the equivalent number of bits there is log2(100^9) = 60. so you'd need about 5 words (5*12=60) for the same level of security.

tl;dr: you need more than 5 words (very) roughly. i am not 100% sure i have the maths right, either :o)

in more practical terms, i don't know of a program that actually does this. the programs i know about typically only combine a few words. but i am no great expert, and it would be fairly easy to write such a program.

[edit: actually, you are less safe than i estimate above if the attacker is smart, because the words probably aren't independent (most people will use meaningful sentences). so you can use markov chains to hugely reduce the search space (in more general terms: "the" will be followed by a noun phrase, etc). so you need significantly more than 5 words. heh. if someone wants to pay me to write such a program, please get in touch... ;o]