# Rate law for A + B → C → P

I need some clarification for the following assignment:

Derive the rate law for

$$\ce{A + B -> C -> P}$$

when $$\ce{A + B -> C}$$ is the slowest step and very slow.

My understanding is $$\ce{A + B -> C}$$ is the rate-determining step, so the rate law would just be

$$\mathrm{rate} = k[\ce{A}][\ce{B}].$$

I'm not sure if the question is asking about steady state approximation (SSA) or just a simple rate law. I know SSA happens if you assume the second step $$(\ce{C -> P})$$ is faster than $$\ce{A + B -> C},$$ meaning $$\ce{A + B -> C}$$ is the slow step. I'm not so sure if my understanding of the question is correct.