Combinatorics problem: form a team of three people with at least one medic, having 5 nurses and 4 medics. At an ambulance stations there are 5 nurses and 4 medics. Find in how many ways can we create a team with three people, where at least one of them is a medic.

I solved the problem this way: there are $4\cdot C_5^2$ ways to form a team with one medic and two nurses and $C_4^2 \cdot 5$ ways to form a team with two medics and one nurse and $C_4^3$ ways to form a team with three medics.

So the answer will be $4\cdot C_5^2 + C_4^2 \cdot 5 + C_4^3 = 74$.

I am not sure if I am right with this answer, so this is the reason I am asking this question.
The quicker way to look at it is that there are $\binom{9}{3}$ ways of selecting a team of 3 people, of which $\binom{5}{3}$ don't have a medic. So the answer is $\binom{9}{3} - \binom{5}{3} = 84 - 10 = 74$, which agrees with your calculation.
Best way

(Count of those with at least 1 medic) = (Count of all combinations) - (Count those 
                                         with no medics)

$${}^9\mathrm C_3-{}^5\mathrm C_3 = 74$$