Disjoint Poisson Arrivals in the Same Time Inteval

To calculate the probability of $$3$$ Poisson arrivals in the interval $$[0,t]$$:

$$(\lambda t)^3 \ e^{-\lambda t} \over 3!$$. However, if we want to answer the question of having two arrivals and one arrival in the same time interval, it will be:

$$(\lambda t)^2 \ e^{-\lambda t} (\lambda t)^1 \ e^{-\lambda t} \over 2!* 1!$$

Which doesn't equal to the above. What am I missing here?

For a Poisson process with rate $$\ \lambda\$$, the quantity $$\ \frac{(\lambda t)^ne^{-\lambda t}}{n!}\$$ is the probability of exactly $$\ n\$$ arrivals occurring in the interval $$\ [0,t]\$$. Your product $$\ \frac{(\lambda t)^3e^{-\lambda t} (\lambda t)^1e^{-\lambda t}}{3!1!}\$$ is therefore the product of the probabilities of exactly $$3$$ arrivals occurring in the interval $$\ [0,t]\$$ and exactly $$1$$ arrival occurring in the interval $$\ [0,t]\$$. But the event that exactly $$3$$ arrivals occur in the interval $$\ [0,t]\$$ and the event that exactly $$1$$ arrival occurs in the interval $$\ [0,t]\$$ are not independent, so the probability of their conjunction is not the product of their probabilities. The two events are in fact disjoint, so the probability of their conjunction is zero.
However, if "having two arrivals and one arrival in the same time interval" is intended to mean "having at least two arrivals and at least one arrival in the same time interval", then the occurrence of the first event implies the occurrence of the latter, so the probability of their conjunction is just the probability of the former, namely $$\ 1- e^{-\lambda t}-\lambda t e^{-\lambda t}\$$.