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If we have 7 rings of different colours and 5 fingers, then what's the number of ways can we wear them? The answer given in the book is $\frac{11!}{4!}$.

The answer is not $7^5$ or $5^7$. They are considering the arrangement of rings in each finger also.

Ottavio
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R banik
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2 Answers2

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Consider the fingers as 5 boxes, and rings are distinctive balls, and orders matter(this is why we cannot use the $5^7$). So
|O(red)O(green)|O(yellow)O(black)|O(pink)|O(white)|O(orange)|
and
|O(green)O(red)|O(yellow)O(black)|O(pink)|O(white)|O(orange)|
are different cases.

This problem could be reduced to: There is 11 positions, 7 for the balls and 4 for the identical walls(the walls in the ends cannot be moved) That means we need to find 4 position to place the walls -- ${11 \choose 4}$ We need to permute the 7 distinctive balls to place them in the remain positions. -- $7!$

So together, it would be ${11 \choose 4}*7! = \frac{11!}{4!}$.

The trick here is the fingers are stacks, so the order of rings on the finger matters.

Flora Sun
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  • Welcome to MathSE. Nicely explained. – N. F. Taussig Jul 13 '20 at 16:17
  • Thank you. :) This is a good place to learn math. – Flora Sun Jul 13 '20 at 16:21
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    If all seven rings must be worn (a reasonable interpretation of the problem), this becomes a "stars and bars" computation, a familiar exercise in elementary combinatorics. See here for the "canonical question" treatment (aka abstract duplicate). – hardmath Jul 13 '20 at 16:26
  • @hardmath The rings are distinct, so that method does not apply. – N. F. Taussig Jul 13 '20 at 17:47
  • @N.F.Taussig: That's a good point, but it is a variation on a theme. Sometimes an exercise involves books on bookshelves, etc. where the items are distinct. If there are $n$ items (as here $n=7$), one has a factor $n!$ that counts all the permutations of these items. – hardmath Jul 13 '20 at 17:51
  • @N.F.Taussig I feel like if I interpret it in the way mentioned in the answer, I do used a little bit of "stars and bars" here to place the walls. However, before I put in all the balls, I permute them because they are different. Then it occurs to me that this question can also be interpret other way -- before I put the first ring, there is 5 available place to go. And because order matters, when I put the second ring, there is 6 (before or after the 1st ring on same finger). so for all 7 rings, there would be 567891011 situations. Then in this way, I do not use the 'stars and bars'. – Flora Sun Jul 13 '20 at 18:55
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At least one ring should be held by the person so u take extra 6 fingers just for reference not real , then no of ways of selecting 7 fingers is $11C_7$ U can see u are at least selecting 1 real finger , now as the rings are different u have arrange them so multiply by 7! Problem solved if u have any doubts u can comment

Namburu Karthik
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  • What was the purpose of taking extra 6 fingers and what do you mean by one ring should be held by the person? – R banik Jul 13 '20 at 16:40
  • Because if at least one finger is also not held by the person what are u doing with those rings – Namburu Karthik Jul 13 '20 at 16:43
  • By the 6 fingers I mean that , u have 7 rings u should at least put one ring to him so this is just a method to do it , by taking extra 6 fingers so at least he will have 1 finger with ring up to 5 – Namburu Karthik Jul 13 '20 at 16:45