8. (10 points) Three households, A, B, and C, plan to subscribe to newspapers. There are 7 different newspapers available. It is known that each household subscribes to 3 different newspapers, and any two households have exactly one newspaper in common. Therefore, the three households have $\qquad$ different subscription methods.
【Solution】Solution: Household A has $\mathrm{C}_{7}^{3}=35$ choices,
Household B needs to choose one of the newspapers that Household A has chosen, and the other two from the newspapers that Household A has not chosen, so there are $\mathrm{C}_{3}^{1}$
$\cdot C_{4}^{2}=18$ choices,
Household C can either choose the newspapers that both A and B have subscribed to, and then choose one from the newspapers that neither A nor B have subscribed to (only 2 left), or choose one from the two different newspapers that A and B have each subscribed to, and then pick one that neither A nor B have chosen, so there are $\mathrm{C}_{2}^{1} \cdot \mathrm{C}_{2}^{1} \cdot$ $C_{2}^{1}+1=9$ choices,

Therefore, there are a total of $35 \times 18 \times 9=5670$ choices,
The answer is 5670.