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Order doesn't matter. Other sites say the answer is $4^9$, which makes sense because each ring has four possible fingers it can be placed on. However, I was wondering why this formula gives me a different answer: $9+4-1\choose 9$ supposedly gives me the distribution of $9$ rings to $4$ fingers. Isn't this the same thing as above?

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    I believe that's the distribution of four fingers over nine rings. Some rings are left out. – 2017-02-01
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    Would swapping the latter 9 with a 4 be the distribution of nine rings over four fingers then? Though I still get a different answer with that, too. – 2017-02-01
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    Not the same; your second formula assumes the rings are identical, so effectively it gives you how many rings are on each finger but not which rings. – 2017-02-01
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    The formula you used is for **identical** rings on distinct fingers, which is not the case here. – 2017-02-01
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    Ah, I see the difference now. Thanks! – 2017-02-01

2 Answers 2

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This is actually a complicated question, since it is possible to put more than one ring into one finger and the order in which they can be put. If we assume we can put all nine rings in one finger:

Case 0: No rings on each finger.

Number of ways: 1.

Case 1: All the rings are in only one finger.

Total number of ways: $4(\binom{9}{1}(1!)+\binom{9}{2}(2!)+...+\binom{9}{9}(9!))=3945636.$

Case 2: All the rings are in only two fingers.

Number of ways to arrange the fingers: $\frac{4!}{2!}=12.$

Case 2a 1+x, where x is between 1-8:

$\binom{9}{1}[\binom{8}{1}(1!)(1!)+\binom{8}{2}(1!)(2!)+...+\binom{8}{8}(1!)(8!)]=986400.$

Case 2b 2+x, where x is between 2-7:

$\binom{9}{2}[\binom{7}{2}(2!)(2!)+\binom{7}{3}(2!)(3!)+...+\binom{7}{7}(2!)(7!)]=985824.$

Case 2c 3+x, where x is between 3-6

$\binom{9}{3}[\binom{6}{3}(3!)(3!)+\binom{6}{4}(3!)(4!)+...+\binom{6}{6}(3!)(6!)]=967680.$

Case 2d 4+x, where x is between 4-5

$\binom{9}{4}[\binom{5}{4}(3!)(3!)+\binom{6}{4}(3!)(4!)+...+\binom{6}{6}(3!)(6!)]=725760.$

Total number of cases: $12(986400+985824+967680+725760)=43987968.$

Case 3: All the rings are in only three fingers.

Number of ways to arrange the fingers: $4!=24$

Case 3a: 1+x+y where x and y are between 1-7:

$\binom{9}{1}[\binom{8}{1}\binom{7}{1}(1!)(1!)(1!)+...+\binom{8}{4}\binom{4}{4}(4!)(4!)]=3239208.$

Case 3b: 2+x+y where x and y are between 2-5:

$\binom{9}{2}[\binom{7}{2}\binom{5}{2}(2!)(2!)(2!)+...+\binom{7}{3}\binom{4}{4}(2!)(3!)(4!)]=1693440.$

Case 3c: 3+3+3:

$\binom{9}{3}[\binom{6}{3}\binom{3}{3}(3!)(3!)(3!)]=362880.$

Total number of ways: $24(3239208+1693440+362880)=127092672.$

Case 4: All the rings are in all four fingers.

Number of ways to arrange the fingers: $4!=24$

Case 4a: 1+x+y+z, where x,y and z are between 1-6:

$\binom{9}{1}[\binom{8}{1}\binom{7}{1}\binom{6}{1}(1!)(1!)(1!)(1!)+...+\binom{8}{2}\binom{6}{3}\binom{6}{3}(1!)(2!)(3!)(3!)]=1270080.$

Case 4b: 2+x+y+z, where x, y and z are between 2,3:

$\binom{9}{2}[\binom{7}{2}\binom{5}{2}\binom{3}{2}(2!)(2!)(2!)(2!)+\binom{7}{2}\binom{5}{2}\binom{3}{3}(2!)(2!)(2!)(3!)]=725760.$

Total number of ways: $24(1270080+725760)=47900160$

Adding up all the cases we have:

$1+3945636+43987968+127092672+47900160=222926437.$

or about 230 million ways.

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Since each of the $9$ rings can go to any of the four finger, there can be $4^9$ ways. This answer assumes that order in which rings are put does not matter.