9. [6] I have four distinct rings that I want to wear on my right hand hand (five distinct fingers.) One of these rings is a Canadian ring that must be worn on a finger by itself, the rest I can arrange however I want. If I have two or more rings on the same finger, then I consider different orders of rings along the same finger to be different arrangements. How many different ways can I wear the rings on my fingers?
Answer: 600 . First we pick the finger for the Canadian ring. This gives a multiplicative factor of 5 . For distributing the remaining 3 rings among 4 fingers, they can either be all on the same finger $(4 \cdot 3$ ! ways), all on different fingers $\left(\binom{4}{3} \cdot 3\right.$ ! ways), or two on one finger and one on another $\left(4 \cdot\binom{3}{2} \cdot 2!\cdot 3\right.$ ways.) Therefore, I have $5 \cdot(24+24+72)=600$ choices.