COMBINATORICS-2

Is this just 4*4*4 --> total 64 ways?

gmat7202011 wrote:In how many ways can 4 rings be worn on 3 fingers (index finger, middle finger and ring finger), if there is no restriction on the number of rings worn on a finger ?

Thank You

should it not be 3^4 = 81 ways

each ring can be worn in 3 ways. so 4 rings can be worn in 3*3*3*3 = 81 ways

@Anurag, 1 finger can hold even 0 OR more than 0 ring(s) - to be precise, as there's no restriction per the number of rings on a finger. Otherwise how one finger can hold 4 rings and yet the other will hold something? It's the same as srss25 says one ring to more than 1 finger ...

Anurag@Gurome wrote:

srcc25anu wrote:should it not be 3^4 = 81 ways

each ring can be worn in 3 ways. so 4 rings can be worn in 3*3*3*3 = 81 ways

Each ring cannot be worn in 3 ways, but in only 1 way.
This is because 1 ring cannot go to more than 1 finger.
But, 1 finger can hold more than 1 ring.
So, a finger can be repeated in 4 ways.
Remember, that which can be repeated goes as the power!

The correct answer is 81.

Thank You all for initiating the discussion, makes me think clearer.

@gmat7202011: WHAT'S THE SOURCE OF THIS QUESTION?
@Anurag, why the solution get's deleted?
@srcc25anu, please explain distribution formula 3^4?

Firstly, is this is any GMAT question? Secondly, the problem does not specify if the rings are identical or not. BUT let's assume both cases for now

Identical rings - ORDER does not matter --> 4C3 * 4C3 *4C3 = 4*4*4=64 ways
Rings are Unique - ORDER does matter --> using the formula n!*C(n+r -1, r-1) we get 4!*6C2 = 360 ways

please reveal the SOURCE and not the "correct" answer

gmat7202011 wrote:In how many ways can 4 rings be worn on 3 fingers (index finger, middle finger and ring finger), if there is no restriction on the number of rings worn on a finger ?

Thank You

Night reader wrote:Is this just 4*4*4 --> total 64 ways?

gmat7202011 wrote:In how many ways can 4 rings be worn on 3 fingers (index finger, middle finger and ring finger), if there is no restriction on the number of rings worn on a finger ?

Thank You

I thought that it will be 4_ * _3 * _2

If identical we have several cases:

4 on one finger- - - - choose one of 3 fingers = 3
3 on one finger, 1 on another---- choose one of 3 fingers to hold 3, another to hold 1 = 6
2 +1 +1- choose one of three fingers to hold 2, the others hold one each = 3
2 + 2 - choose finger to hold none, the others hold two each= 3

Thus 15


If distinct, each of the 4 rings can go on any of the three fingers = 3^4 = 81

Hi Kevin,

Thank for the response, I guess, we need to consider them distinct rings, since there is no mention of they being identical.

Night Reader,

I ran into this question from a Math Quant book by Arun sharma, i found in a library in Chicago.

Also, "reveal" is a word you use, when you hide something, i guess a better phrase in this context would have been
"Can you please let us know the source of the question ?"


"reveal the source" not the correct answer, i guess mentioning the correct answer is important too.

Thank You

A GMAT question would probably specify whether they were different or identical, but I thought it would be nice to present a solution for each case. It is, however, nice to disclose/reveal/mention/cite the source, as some people may wish to avoid doing GMATPrep questions until they have used the simulation software by the same name several times

kevincanspain wrote:If identical we have several cases:

4 on one finger- - - - choose one of 3 fingers = 3
3 on one finger, 1 on another---- choose one of 3 fingers to hold 3, another to hold 1 = 6
2 +1 +1- choose one of three fingers to hold 2, the others hold one each = 3
2 + 2 - choose finger to hold none, the others hold two each= 3

Thus 15


If distinct, each of the 4 rings can go on any of the three fingers = 3^4 = 81

Perfect kevincanspain !

For the identical case, we can also look at the problem in another way which reduces the calculation. We want to distribute 4 things (4 rings) among 3 people (3 fingers here), where each one can receive 0 or more things =
(n+r-1)C(r-1) = (4+3-1)C(3-1)
= 6C2 = 15.