It's a set problem. There are 6 houses, those are our sets. We have 9 pets of three types (elements of our sets). There are finite ways to put these elements into sets:

∅ (empty set - no pets)
{R} (one rabbit)
{D} (one dog)
{C} (one cat)
{R, C} (one rabbit, one cat)
{R, D} (etc.)
{D, C}
{R, D, C}

From the problem we can eliminate ∅, {D, C}, {R, C}, and {R, D, C}. So houses can have either one or more of a single animal, or dog(s) and rabbit(s). All the combinations I can see require at least one house with at least one rabbit and dog. The rest is how you distribute the remaining pets according to those rules (eg the cats will always be alone or with other cats).

There is a more formulaic and confusing way to model this problem but it's really quite complicated, or maybe it's just late and my brain's not in gear. I could calculate the number of possible combinations from this point but I have a degree in maths... I agree with other comments suggesting that the intention was more to encourage kiddo to figure out these rules and apply trial and error - it's not really a combinatorics problem in the sense that they want THE provable answer but they want the kids to a) notice that the houses are the sets or "buckets" that items are being placed into and b) try and distribute the pets into the houses and recognise when they've broken one of the descriptions.

Really impressed with kiddo for taking to investigating it. The mathsy maths used to actually solve the problem is something I learned in uni, but so are the basic concepts this is prompting. And I was like five times as old as your kid! Don't feel dumb for not being able to do any better, this is a tough problem and doesn't involve the sort of maths taught in school. The only difference between a trial and error method and the calculable solution is that with the former you won't know for sure when you're done and have gotten all the combinations. That's fine, they just asked to find as many as you can.