We want to distribute $6$ people $ P_1, P_2, P_3, P_4, P_5 $ and $ P_6 $ to $4$ houses $ H_1, H_2, H_3 $ and $ H_4 $. Each person chooses, which house to go. We are searching for the probability that two people are found alone, and the other four in two pairs.
Firstly, I found that there are totally $ 4^6 = 4096 $ contingencies, as all $ 6 $ people are free to choose among $ 4 $ houses.
Secondly, the number of contingencies that two people are alone, each at a house, and the other four in two pairs, in the two remaining houses, should be :
$$ 4! { 6 \choose 2 \; 2 \; 1 \; 1 } = \cfrac{4!6!}{2!2!1!1!} = 4320 $$
Apparently, something is wrong, as the requested probability will be greater than $ 1 $, as $ 4320 > 4096 $, but I cannot find the mistake.
Thank you in advance.
There are $$\dfrac{4!}{2!2!}$$ ways of ordering the houses so that you have two houses with 2 occupants and two houses with 1 occupant. There are $$\dfrac{6!}{2!2!1!1!}$$ ways of ordering the people into those houses. So, you forgot to divide by 4.
Total:
$$\dfrac{4!6!}{2!2!2!2!1!1!} = 1080$$