How many distinct factors can be made from the number $2^5*3^4*5^3*7^2*11^1$? How many distinct factors can be made from the number $2^5*3^4*5^3*7^2*11^1$?
Hmmm... So I didn't know what to do here so I tested some cases for a rule. 
If a number had the factors $3^2$ and $2^1$, you can make $5$ distinct factors: $2^1$, $3^1$, $3^2$, $2^1 \cdot 3^1$, $2^1 \cdot 3^2$... I don't see a pattern yet.
How does one go about this?
I don't think the answer is $5!$....

Your approach with $3^2$ and $2^1$ is very close. You only forgot to account for the factor $1 = 3^0\cdot2^0$. This brings us to $6$ distinct factors for this smaller example. You can calculate this by multiplying the number of all the different exponents used for each factor. $3$ can occur either $2$, $1$ or $0$ times, while $2$ can only occur $1$ or $0$ times. This brings us to $3\cdot2 = 6$ combinations of these exponents.  
Now, the "real" question has another little trick to it because the factors aren't all that nice just yet. $2^5\cdot3^4\cdot4^3\cdot3^2\cdot2^1 = 2^6\cdot3^6\cdot4^3$
Much better.
Now, we can have between $0$ and $6$ occurences of $2$, making it $7$ possibilities.
Another $7$ possibilities for occurences of $3$ and finally $4$ possibilities for $4$.
This makes a total of $7\cdot7\cdot4 = 196$ possible distinct factors of $2^6\cdot3^6\cdot4^3 = 2^5\cdot3^4\cdot4^3\cdot3^2\cdot2^1$.
Edit
Alright, I didn't pay attention either. As Konstantinos Gaitanas stated, $4^3 = 2^6$, making the whole thing $2^{12}\cdot3^6$.
So the solution should indeed be $13\cdot7$ = 91.
