My thinking behind this problem would be to pick $2$ digits out of 10 total $\binom{10}{2}$, $4$ consonants out of $21$, vowels not included $\binom{21}{4}$, and the $5$ vowels, multiplied by $7!$.
$\binom{10}{2}*\binom{21}{4}*5*7!$
I am unsure what to divide by though to get an actual probability! $36!$ possibly?
On
Assuming that you're just counting the latin alphabet (not differentiating between lower and upper), and not allowing any other characters in the string, then the total number of possible strings is $36^7$ if you allow repetition and 36 permutation 7 if you don't. This is equal to $36.35.34.33.32.31.30=\frac{36!}{(36-7)!}=\frac{36!}{29!}$
The number of strings of length $7$ made from digits and letters with no repetition is $36!/(36-7)!=36*35*\cdots*30$, but that's taking order into account. If you take order into account in the denominator, you have to do so in the numerator as well. Since we're given that there's no repetition, each particular choice of $7$ characters for our string could happen in $7!$ different orders, so we can ignore order without causing problems.
Thus, ignoring order, there are $\binom{36}{7}$ sets of characters that our string could be composed of. Your count for the numerator is right, but you can drop the $7!$. If you keep the $7!$ in the numerator, then use $\frac{36!}{29!}$ for your denominator.
Thus, your answer is:
$$\frac{\binom{10}{2}\binom{21}{4}\binom{5}{1}7!}{\frac{36!}{29!}}=\frac{\binom{10}{2}\binom{21}{4}\binom{5}{1}}{\binom{36}{7}}=\frac{29925}{185504}\approx 16.13\%$$