jmc

Place Fluffy in the 3-dog group and Nipper in the 5-dog group. This leaves 8 dogs remaining to put in the last two spots of Fluffy's group, which can be done in $\left(\genfrac{}{}{0ex}{}{8}{2}\right)$ ways. Then there are 6 dogs remaining for the last 4 spots in Nipper's group, which can be done in $\left(\genfrac{}{}{0ex}{}{6}{4}\right)$ ways. The remaining 2-dog group takes the last 2 dogs. So the total number of possibilities is $\left(\genfrac{}{}{0ex}{}{8}{2}\right)\times \left(\genfrac{}{}{0ex}{}{6}{4}\right)=\boxed{420}$.