Actually, both can be correct,although $9 \times 2^4 +1$ makes more sense.
One thing that you should realize (and no, this is not told usually in class, so it is important) is that there is no problem if you include "extra" outcomes in your sample space which cannot occur. You can simply adjust these to have probability zero, so that in any calculations the "extra" outcomes will not be considered because they will have probability zero.
For example, if you do include the outcome that you took the biased coin and got $4$ heads, say, then this could occur in the sample space : but it would have probability zero. This way, you could have $10 \times 2^4$ outcomes, but some of them would be extra, so will be given probability zero.
On the other hand, you could choose not to include these at all. This would lead to $9 \times 2^4$ outcomes for each of the fair coins, and $1$ for the biased coin, giving your second answer. This is more trimmed, but I would accept the first as an answer if the probabilistic explanation accompanied it.