Consider the following state space: $$ \Omega := \{\omega : \omega = (a_1,a_2,a_3,a_4,a_5), a_j \neq a_k, j\neq k, a_i \in \{1,\dots,52\}\}$$
It models sampling without replacement and can be used for calculating probabilities of poker hands containing $5$ cards of a standard $52$ cards deck. Now I want to calculate the probability of a full house, i.e. three cards of one rank and two cards of another rank. So let us consider the case where the three cards of one rank come first and then the two others. For the first card we have $52$ possibilities and then $3$ and $2$ for the second and third since there are only $4$ cards of one rank. Now for the fourth we have $48$ possibilities since it should be not of the same rank as the previous one. Finally, for the fifth one we have $3$ possibilities. So multiplying this by $5!$ yields the desired possibilities and so we get $$P(\{\text{full house}\}) = \frac{5! \cdot 52 \cdot 3 \cdot 2 \cdot 48 \cdot 3}{52 \cdot 51 \cdot 50 \cdot 49 \cdot 48} \approx 0.017$$ but according to here we should get $$P(\{\text{full house}\}) \approx 0.0014$$ What is wrong with my reasoning? It worked for the previous exercises but somehow not here.