The number of eligible boys is $b = 12$.
We need to choose 2 girls as captain and vice-captain, and 2 boys as captain and vice-captain.
First, we choose 2 girls from 3 eligible girls. The number of ways to do this is given by the number of permutations of choosing 2 girls from 3, which is $P(3, 2) = \frac{3!}{(3-2)!} = 3 \times 2 = 6$.
The order matters here since one is the captain and the other is the vice-captain.
Next, we choose 2 boys from 12 eligible boys. The number of ways to do this is given by the number of permutations of choosing 2 boys from 12, which is $P(12, 2) = \frac{12!}{(12-2)!} = 12 \times 11 = 132$.
The order matters here since one is the captain and the other is the vice-captain.
To find the total number of ways to choose the captains and vice-captains, we multiply the number of ways to choose the girls by the number of ways to choose the boys:
Total number of ways = (Number of ways to choose 2 girls) $\times$ (Number of ways to choose 2 boys)
Total number of ways = $6 \times 132 = 792$
Therefore, there are 792 ways to choose four students, two girls as captain and vice-captain and two boys as captain and vice-captain.
Final Answer: The final answer is $\boxed{792}$