Library / Annual Harvard-MIT Mathematics Tournament /14 of 72

Annual Harvard-MIT Mathematics Tournament

Combinatorics Difficulty 4.7 AIME Find the answer United States

Problem:

I have four distinct rings that I want to wear on my right 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?

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Solution:

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 ( $\binom{4}{3} \cdot 3!$ ways), or two on one finger and one on another ( $4 \cdot \binom{3}{2} \cdot 2! \cdot 3$ ways). Therefore, I have $5 \cdot (24 + 24 + 72) = 600$ choices.

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.