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Re: 5 rings on 4 fingers. [#permalink]
Answers
1) A

2) C

3) C

OA pls
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Re: 5 rings on 4 fingers. [#permalink]
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Bunuel wrote:
Saw the question below (#1) in a GMAT tough math problems. Think that the solution given is not correct. It's a good question to train in combinatorics so give it a try:

1. In how many ways can 5 rings be worn on the four particular fingers of the right hand?
(A) 4^5
(B) 5^4
(C) 8C3
(D) 8P3
(E) 4!*5!


1st ring can be worn on any of the 4 fingers = 4 ways
2nd ring can be worn on any of the 4 fingers = 4 ways
3rd ring can be worn on any of the 4 fingers = 4 ways
4th ring can be worn on any of the 4 fingers = 4 ways
5th ring can be worn on any of the 4 fingers = 4 ways

Total possibilities = 4 x 4 x 4 x 4 x 4 = 4^5
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Re: 5 rings on 4 fingers. [#permalink]
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GMAT TIGER wrote:
Bunuel wrote:
Saw the question below (#1) in a GMAT tough math problems. Think that the solution given is not correct. It's a good question to train in combinatorics so give it a try:

1. In how many ways can 5 rings be worn on the four particular fingers of the right hand?
(A) 4^5
(B) 5^4
(C) 8C3
(D) 8P3
(E) 4!*5!


1st ring can be worn on any of the 4 fingers = 4 ways
2nd ring can be worn on any of the 4 fingers = 4 ways
3rd ring can be worn on any of the 4 fingers = 4 ways
4th ring can be worn on any of the 4 fingers = 4 ways
5th ring can be worn on any of the 4 fingers = 4 ways

Total possibilities = 4 x 4 x 4 x 4 x 4 = 4^5


Well this was the answer given in the source. But I don't agree. Consider the following:

We have 3 rings and 2 fingers. According to this explanation the answer would be 2^3=8. But let's count:
30
03
21
12
Only 4 possible ways, so 2^3 and 4^5 in our original question have duplications. That's why I posted this question, I found the answer and explanation given are wrong.

My answer and solution is totally different.

So what do you think?
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Re: 5 rings on 4 fingers. [#permalink]
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smtripathi wrote:
I think we can solve it like :

Conside 4 fingers

I II III IV

5 0 0 0
4 1 0 0
3 2 0 0
2 2 1 0
1 2 1 1

All the cases above can each be arranged in 4 ways.
1*4+ (5c4*5c1)*4+(5c3*5c2)*4+(5c2*5c2*5c1)*4+(5c1*5c2*5c1*5c1)*4


This would be very large number, even larger than 4^5 and will definitely have duplicates.
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Re: 5 rings on 4 fingers. [#permalink]
I am not sure if I am right.. And I have given the answer in numerical terms..

I would be happy if you can confirm if atleast the approach is right..

My line of thinking goes like this..


For the first question, (before reading the second question), we are lead to believe that the rings are not identical. But after reading the second question, if we assume that the all the rings in the first question are identical, then I think there is just one way of wearing 5 rings in the four particular fingers of right hand.

So the ans is 1

(since the fingers are particular we dont get combinations there, and since the rings are identical we dont get combinations there also)

For the second question, given the rings are different and we have 4 particular fingers, it goes like
for the 1st ring u ve got 4 free fingers
for the 2nd ring u ve got 3
for the 3rd ring u ve got 2
for the 4th ring u ve got 1

4X3X2X1 = 24 possiblities. BUT, there is one ring left out. This means that for ever ring left out there aer 24 possibilties.

In other words If R1 is selected first there are 24 possibilities, if R2 is selected first there are 24 possibiltes and so on..

So ans is 24x5 = 120

For the third question we are free to choose any 4 of the 5 fingers.. So, it just the matter of multiplying the ans for the second question with the no. of finger combinations.
No of possible finger combinations are 5! / (4!x1!) = 5

So the ans is 120x5 = 600

ny comments guys ??

Bunuel wrote:
No one else wants to try? I don't have OA for these questions (well I have for the first one but as I said I disagree) so can only post my answers if no more attempts.
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Re: 5 rings on 4 fingers. [#permalink]
All look so confusing and are debatable.
Please guys share your views.
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Re: 5 rings on 4 fingers. [#permalink]
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