5 rings on 4 fingers.

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5 rings on 4 fingers. [#permalink]

31 Oct 2009, 23:35

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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!

2. In how many ways can 5 different rings be worn on the four particular fingers of the right hand?
(A) 4^5*5C4
(B) 5^4*5C4
(C) 8C3
(D) 8P3
(E) 4!*5!

3. In how many ways can 5 different rings be worn on the any four fingers of the right hand?
(A) 4^5*5C4
(B) 5^4*5C4
(C) 8C3*5C4
(D) 8P3*5C4
(E) 4!*5!*5C4
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Re: 5 rings on 4 fingers. [#permalink]

01 Nov 2009, 12:27

These are real brain teasers... I would go with option E in all 3. Not quite certain though.. will wait for answers from others..

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Re: 5 rings on 4 fingers. [#permalink]

01 Nov 2009, 21:33

Answers
1) A

2) C

3) C

OA pls

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Re: 5 rings on 4 fingers. [#permalink]

02 Nov 2009, 00:13

Bunuel wrote:

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]

04 Nov 2009, 16:48

GMAT TIGER wrote:

Bunuel wrote:

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]

05 Nov 2009, 09:56

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]

09 Nov 2009, 03:15

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]

09 Nov 2009, 04:45

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I think it's time to post my own solutions:

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!

OA for this question is 4^5 with OE as Economist showed in his post. But I do think that it's not correct, see my reasoning of it in my previous post.

We have 5 identical rings and 4 particular fingers. We want to count how in how many ways these 5 rings can be worn on 4 fingers. Basically it's the same as the total number of ways of dividing n identical items among r persons, each one of whom, can receive 0,1,2 or more items, which is:
n+r-1Cr-1.

In our case n=5 and r=4: (5+4-1)C(4-1)=8C3=56

Answer: C.

2. In how many ways can 5 different rings be worn on the four particular fingers of the right hand?
(A) 4^5*5C4
(B) 5^4*5C4
(C) 8C3
(D) 8P3
(E) 4!*5!

This case says that rings are not identical, as n rings on the particular finger can be worn in n! ways, we should multiply 8C3 by 5! which is 8!/3!=6720. We don't have this answer choice. But I really think this should be the answer.

Consider this:
First ring can be worn in 4 ways;
Second ring can be worn in 5 ways (as it can gon on any of four fingers - 4 ways; plus it can go below the first one - 1);
Third ring can be worn in 6 ways (the same as for the second);
Fourth ring can be worn in 7 ways;
Fifth ring can be worn in 8 ways;

Total: 4*5*6*7*8=6720.

The answer choice D should be 8P5 instead of 8P3.

3. In how many ways can 5 different rings be worn on the any four fingers of the right hand?
(A) 4^5*5C4
(B) 5^4*5C4
(C) 8C3*5C4
(D) 8P3*5C4
(E) 4!*5!*5C4

Here we should just multiply previous answer by 5C4 the number of selection of four fingers out of five: 6720*5C4=6720*5=33600.

The answer choice should be 8P5*5C4 instead of 8P3*5C4.
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Re: 5 rings on 4 fingers. [#permalink]

16 Feb 2010, 11:32

All look so confusing and are debatable.
Please guys share your views.
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Re: 5 rings on 4 fingers. [#permalink] 16 Feb 2010, 11:32

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5 rings on 4 fingers.

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