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In how many ways $5$ different rings can be worn on $4$ fingers ?

Although these is similar question here In how many ways $3$ different rings can be worn in $4$ fingers with at most one in each finger?

But I want to answer it in different way which is like take $4$ fingers like $a,b,c,d$. Now $a$ can be filled in $5$ ways and so are others. So total ways are $5^4$ but answer is $4^5$. What is wrong in my reasoning? Which cases have I left out?

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  • $\begingroup$ If you fill $a$ with $5$ then b, c, d don't have any rings. $\endgroup$ – kingW3 Apr 7 '17 at 11:55
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    $\begingroup$ The answer is $4^5$ because for each of the rings, you have four fingers to choose from. $\endgroup$ – Lazy Lee Apr 7 '17 at 11:55
  • $\begingroup$ @LazyLee why can't we have for each of 4 fingers , you have 5 rings to choose from $\endgroup$ – Taylor Ted Apr 7 '17 at 11:56
  • $\begingroup$ @kingW3 can you elaborate? $\endgroup$ – Taylor Ted Apr 7 '17 at 11:59
  • $\begingroup$ No, because for each of the $5$ rings, you have to select one of the fingers to put it on. Your case is for another problem: "How many ways can you put one ring onto each of four fingers, when each of the fingers can choose from 5 rings". $\endgroup$ – Lazy Lee Apr 7 '17 at 11:59
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The answer is $4^5$ because you have $4$ fingers for each of the ring. Or, in other words each ring have $4$ choices. $$4\times4\times4\times4\times4=4^5$$

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I'm assuming that the four fingers are given, e.g., the fingers of one hand without the thumb. An arrangement of five distinguishable rings on the four labeled fingers amounts to a linear arrangement of $1$, $2$, $3$, $4$, $5$, and three indistinguishable zeros as separators. There are ${8!\over3!}=6720$ such arrangements. Note that a green ring and a blue ring on the index finger can be worn in $2$ ways.

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  • $\begingroup$ If we let $0$ denote the lack of a ring and if given a linear arrangement, the first pair go on the first finger (in the specified order), etc then the sequence $10234500$ and $01234500$ should be considered equivalent but I don't see how your solution accounts for that? $\endgroup$ – benguin Apr 9 '17 at 4:07
  • $\begingroup$ @benguin : The zeros serve as separators. $\endgroup$ – Christian Blatter Apr 9 '17 at 7:05
  • $\begingroup$ Thanks for the clarification, that makes more sense :) I was considering an argument that first orders the rings ($5!$ ways) then uses stars-and-bars to place the rings onto the fingers ($C(5+4-1,4-1)$ ways) for a total of $5!C(8,3)$ which gives the same result. $\endgroup$ – benguin Apr 9 '17 at 8:43
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Suppose you had only one ring. It then would be $4^1=4$ different arrangements/placements. Now if we add one more ring, it will be $4\cdot4$ because we combine the four possible arrangements with yet another four arrangements which makes it $16$. This means you can have two rings on one finger. By the same token we would have $4^5=1024$ arrangements for five rings. And the person can have five rings on one finger or no ring at all on the finger.

If you had $5^4$ arrangements, it would mean $5^1=5$ different arrangements for one ring on four fingers. And that doesn’t make sense.

And finger $a$ can have six possible arrangements: either of five rings or no ring at all.

Hope my explanation was not confusing.

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  • $\begingroup$ The answer $4^5$ is correct if we only care about which ring appears on which finger. To see how to handle the problem in which the order of the rings on the fingers matters, see the answer by Christian Blatter. $\endgroup$ – N. F. Taussig Apr 7 '17 at 15:44

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