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smcgrath12
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smcgrath12
BG
guess in 4^5 ways.

Your answer is correct. I got 5x4x3x2. The way I figure is, the 1st finger can have any of the 5 rings. So 5 ways for 1st finger. The 2nd finger can have any of the 4 remaining rings. So 4 ways for 2nd finger and .....



Why is my answer incorrect?


You forget that any of 4 fingers can have all the rings.

The way it works here, take one ring, on how many fingers it can be worn? - 4
take another ring - 4, and so on
so you get 4*4*4*4*4 = 4^5
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The number of ways that four rings can be picked from a set of 5 is 5C4
The number of ways these rings can be worn is 4!

Thus my answer = 5C4 * 4!
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sparky
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cloudz9
The number of ways that four rings can be picked from a set of 5 is 5C4
The number of ways these rings can be worn is 4!

Thus my answer = 5C4 * 4!


but you assume one ring per finger. what if one wears two rings on one finger?
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sparky
cloudz9
The number of ways that four rings can be picked from a set of 5 is 5C4
The number of ways these rings can be worn is 4!

Thus my answer = 5C4 * 4!

but you assume one ring per finger. what if one wears two rings on one finger?


True!
didn't think of that!
thnx :)
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HowManyToGo
BG
guess in 4^5 ways.

I agree with your reasoning, but then woudn't you have to permute the rings on a finger if you put more than one ring on it ( after all , you can put two rings on a finger in two different ways !).

HMTG.


Here you assume all rings are the same, so orders don't matter.
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HowManyToGo
BG
guess in 4^5 ways.

I agree with your reasoning, but then woudn't you have to permute the rings on a finger if you put more than one ring on it ( after all , you can put two rings on a finger in two different ways !).

HMTG.

Here you assume all rings are the same, so orders don't matter.


In which case( all rings the same) , choosing the 2nd ring and the 4th ring on the 1 st finger is the same as choosing the 3rd and 5th ring for the same.
while computing 4^5 this is taken as two distinct ways.

Hence if all the rings are same,I think the no. of ways is

8c3 = 56 ways.

HMTG.
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"the four fingers" sounds like you are talking about 4 specific fingers. Are we considering all 5... or do we only have 4?



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