I feel like I'm overthinking this as the only online example I've come across is this paper, which is using more advanced math than is expected of us. But if we have a known passface database off 100 faces, a password is five faces, and verification is choosing 1 face from a random 9, 5 times. The set size can't be 100 though can it, since in each challenge there's only 9 options. It's not the same 9 though so I don't think log2(9^5) or log2(100^5) are correct.
If you have a set of 100 total unique faces and have to choose 5 different faces to represent your passface, then the total number of possible face combinations is 100 x 99 x 98 x 97 x 96, or 9,034,502,400. Similar to your 100^5 estimate, but slightly less because you can't choose the same face more than once. Some people would say this represents the possible entropy (e.g. 33 bits) of a passface in this scenario.
But that doesn't seem to be what you're asking. You seem more interested in the probability of an attacker randomly guessing a passface using the normal authentication system. If 9 faces are presented for selection then the attacker has a 1 in 9 chance of guessing the correct face. If they have to choose the correct face from the 9 different presented face choices 5 times in a row then the number of possible choices are the 9^5 you mention, or 59,049.
I'm not sure what you mean by "it's not the same 9 [faces] though". If you mean each of the 5 grids don't contain the same 'decoy' faces along with 1 valid face then you're correct. Each of the 5 grids contain a different selection of decoy faces. But the 9^5 possibilities figure still works. You can comment or edit your question if you meant something else.