Permutation problem 2 spaniels, 2 retrievers and 3 poodles go through to the final. They are placed in line. How many different arrangements of these 7 dogs are there if no poodle is next to another poodle. My attempt, spaniels and retrievers are 4!. Then ? I'm stuck here.

Write down four stars like this: $$\ast\qquad\ast\qquad\ast\qquad\ast$$ These represent the positions to be ultimately occupied by the non-poodles.

There are $5$ gaps, the $3$ obvious ones and the $2$ endgaps. We must choose $3$ of these to be occupied by the poodles. This can be done in $\binom{5}{3}$ ways. The individual poodles can be put in the chosen gaps in $3!$ orders, and the rest of the dogs can occupy the starred positions in $4!$ ways, for a total of $\binom{5}{3}3!4!$.