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This is a variation of the problem in the link below, while the original assumed that you could put more than one rings on one finger. I was wondering if we were to have 8 different types of rings and 3 fingers, how many ways can the rings be placed on the fingers if there are no more than 1 ring per finger?

I have a proposed idea on how to solve this problem though I doubt if my approach is correct. Unlike the original problem, the stars and bars method won't work. We have to make the problem divided into mutually exclusive cases, and I was thinking of doing the following:

Case 1: 1 ring ${8 \choose 1}{3 \choose 1}$

Case 2: 2 rings ${8 \choose 2}{3 \choose 2}$

Case 3: 3 rings: ${8 \choose 3}3!$

But my real trouble was not the ${8 \choose x}$, where $x$ is the number of rings, rather it was ${3 \choose y}$ where $y$ is the number of fingers. To clarify, I have doubts because I was wondering that I was going to have to multiply by $3!$ for every single one instead because there were 6 ways of arranging three fingers?

The Original Problem

jvdhooft
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John Rawls
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  • What?how can you arrange fingers? Index finger will be in its place no matter which ring you put. Unless the person has prosthetic fingers which can be removable. – Archis Welankar May 24 '17 at 04:16
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    @ArchisWelankar I am trying to say that putting a diamond ring on a thumb is different from a pinkie – John Rawls May 24 '17 at 04:24

2 Answers2

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You are very close. There are indeed three cases:

  1. One ring: choose 1 ring out of 8 and 1 finger out of 3, with $1!$ ways in which the rings can be placed, for a total of ${8 \choose 1} \cdot {3 \choose 1} \cdot 1! = 24$ possibilities.

  2. Two rings: choose 2 rings out of 8 and 2 fingers out of 3, with $2!$ ways in which the rings can be placed, for a total of ${8 \choose 2} \cdot {3 \choose 2} \cdot 2! = 168$ possibilities.

  3. Three rings: choose 3 rings out of 8 and 3 fingers out of 3, with $3!$ ways in which the rings can be placed, for a total of ${8 \choose 3} \cdot {3 \choose 3} \cdot 3! = 336$ possibilities.

In total, there are $24 + 168 + 336 = 528$ possibilities. Note that if it is also allowed to select no rings at all, there are $529$ possible scenarios.

jvdhooft
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  • the answer is fine but i do believe that you are missing the case in which there are 0 rings – John Rawls May 24 '17 at 04:25
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    @JohnRawls From your question and the three suggested cases, I assumed at least one ring had to be selected. If it is allowed to select no rings at all, this adds to one additonal possibility. I've updated my answer to account for this. – jvdhooft May 24 '17 at 04:26
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Pretty close, but you have to count the same things each case. (And include the fourth case of: ways to put zero rings on the fingers.)

It is:$$\require{cancel}{\quad\cancelto{1}{\binom 80\binom 30~0!}+\binom 8 1 \binom 3 1~ 1!+\binom 82 \binom 32 ~2! +\binom 83\binom 33 ~3!~\\=~529}$$

Counting for selecting 0,1,2, or 3 from 8 rings, selecting that number from 3 fingers, and arrange those selected rings on those selected fingers.

Graham Kemp
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