Maths Olympiad Prep

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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?

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

Solution

Solution:

Answer: 600600. First we pick the finger for the Canadian ring. This gives a multiplicative factor of 55. For distributing the remaining 33 rings among 44 fingers, they can either be all on the same finger (4⋅3!4 \cdot 3! ways), all on different fingers ((43)⋅3!\binom{4}{3} \cdot 3! ways), or two on one finger and one on another (4⋅(32)⋅2!⋅34 \cdot \binom{3}{2} \cdot 2! \cdot 3 ways). Therefore, I have 5⋅(24+24+72)=6005 \cdot (24 + 24 + 72) = 600 choices.

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