Zeynep (the lady of the rings) has 10 identical gold, 10 identical sil

Dear leve,
I'm happy to respond.

My friend, I think I might be beginning to understand the confusion. If I understand correctly, what you are saying is that the final arrangement of the rings once all 30 are on her fingers is completely fixed. For the final arrangement, there is only one possibility, but that's not what you are asking. If I understand correctly, you intend to ask not
In how many different orders can she wear her all 30 rings on her 10 (distinguishable) fingers?
but rather
In how many ways can she execute the processing of putting these 30 rings on her fingers, so that she ends up with this fixed outcome?
Is this what the question is supposed to be asking? If it is, do you understand that this meaning was entirely opaque from your previous formulations? You don't want the number of final arrangements, but instead the number of different ways to get to that single final arrangement. The verb "wear" implied the finished state of being dressed: thus, I have only one way to wear two identical socks on my two feet. By contrast, the verb "putting on" implies the process of getting dressed: thus, I have two ways to put on two socks, because I could start with either foot.

My friend, I gather that you encountered this problem in the past but don't have the exact wording, so you are trying to state the problem in your own wording. I think you are underestimating the high level of precision needed to formulate a Quantitative question in a rigorously unambiguous fashion. Never underestimate the tremendous value of the exact wording of a problem from a reputable source, because the slightest casual paraphrase can unintentionally compromise the logic of the entire question.

I will forebear offering any solution until I am perfectly clear about what is being asked.

Does all this make sense?
Mike

Hi again Mike,

Yes, the question is asking 'In how many ways can she execute the processing of putting these 30 rings on her 10 (distinguishable) fingers, so that she ends up with this fixed outcome? '

I revised the question as you suggested. I believe this time it is okay. Would you please check it?

Dear leve
My friend, the wording of the question is now clear. I don't know whether this is exactly as the GMAT would frame it, but at least it is clear what is being asked at this point.

The only other suggestion I have would be to make the answer choices a little more fancy looking:

A) \(\frac{(10!)^3}{2^{10}}\)

B) \(\frac{(10!)^3}{3^{10}}\)

C) \(\frac{30!}{2^{10}}\)

D) \(\frac{30!}{3^{10}}\)

E) \(\frac{30!}{6^{10}}\)

By convention, the answer choice letters are typically listed as capital letters, not lower case. Also, professional math notation is always preferable to plaintext shortcuts.

Finally, I will say, as intriguing a mathematical question as this is, it is strangely un-GMAT-like. You see, the business world is all about results. Once a result has been achieved, the route one took to achieving it rarely merits much attention. This counting problem is all about the different evanescent choices toward a fixed outcome: do you see how, in this way, it flies in the face of the priorities of the business world, however interesting it may be?

Mike

Hi Mike,

Thanks for your help. You are right about the business world. But we can consider this problem as a high level challenge problem. Since we are happy with the wording of the problem -If it is possible- can you erase the 'poor quality' tag from the tag list and unlock the question? I think this is a good challenging problem for especially advanced combinatorics practice.

The topic is unlocked. Because the wording is not GMAT-like the tags remain.

Okay..this was a really tough one for me..it took me half an hour to figure this one out.

Here it goes..

Let's first list out what is given in the question..

We have 30 rings in total, and 10 fingers on which we have to wear them.
There are 10 Gold, 10 Silver, and 10 Bronze rings. Every finger will have 3 rings worn in a fixed order, i.e. Gold, then Silver above it, and then Bronze on the top.
We have been asked about the number of ways in which we can wear the rings like this.
The question has given hints about the how the 'interactions' or 'movements' will happen. One of the example order of interactions that has been given to us is :-

Example of a possible order: a gold ring to the left index finger, then another gold ring to the right little finger, then another gold ring to the right middle finger, then a silver ring to left index finger, then a bronze ring to left index finger... and so on.

To wear the 30 rings, we will have in total 30 movements of wearing some or the other ring on some or the other finger at a given point of time.
The constraint that we have is - For each finger, Gold has to be worn first, then Silver, and finally Bronze. So we may fully cover any finger in however many movements, but this is how we have to wear them.

I will give 2 more ways in which we can decide the order to wear.

1. Gold-left thumb, Gold-right index finger, Silver-right index finger, Gold-left ring finger, Bronze-right index finger,Gold-right pinkie...and so on

2. Gold-right middle finger, Silver-right middle finger, Gold-left index finger, Gold-right pinkie, Silver-left index finger, Bronze right middle finger...and so on

All ways will have a total of 30 interactions.

Now we're gonna name every interaction and see.

I'm going to call wearing a Gold ring on the 1st finger as \(G_1\), wearing a Silver ring on the 1st finger as \(S_1\), and wearing a Bronze ring on the 1st finger as \(B_1\), and so on for all the 10 unique fingers.

I'm going to now write our 30 interactions in this subscript manner..
\(G_1\),\(G_7\),\(S_1\),\(G_5\),\(S_7\),\(B_1\),.....and so on

as can be seen, each of these interactions is unique.
So, in how many ways can we arrange 30 unique interactions?

\(30!\)

But we have a problem here, for every finger(let's say for 3rd finger)..
\(G_3\) has to come before \(S_3\), and \(S_3\) has to come before \(B_3\)

So this has to be our order \(G_3\)\(S_3\)\(B_3\)

And this has to be true for all the fingers. What I mean to say is, for any of the \(30!\) arrangements, and for every finger, \(G_n\)\(S_n\)\(B_n\) has to be the order of wearing rings. These interactions may be separated by any number of interactions of other fingers, but this has to be the order.

We know that in any of the \(30!\) arrangements, there will be interactions for each finger arranged in some order separated by any number of interactions..
for example..
\(G_n\)\(S_n\)\(B_n\)
\(S_n\)\(G_n\)\(B_n\)
\(B_n\)\(S_n\)\(G_n\)...
\(3!\) = 6 ways.

But we have to count only one of these orders
\(G_n\)\(S_n\)\(B_n\)

This has to be true for all the 10 fingers. So we have to divide \(30!\) by \(6\) for each finger to remove over counting.

Final Answer

\(\frac{30!}{6^{10}}\)

E.

mikemcgarry Bunuel

can something of this sort be expected on the GMAT?

The question is tagged as Out of scope/Too hard, so no. Also, as Mike mentioned above, the wording is not as precise as GMAT uses.

Here's my answer: E

Note that to wear 30 rings she needs to choose her fingers 30 times (once for each ring) - each finger thrice (once for G, once for S and once for B).
Now, she has, F1, F2, F3,.., F10 fingers.
One possibility is that the lady wears G-S-B in F1, then G-S-B on F2, ... , then G-S-B on F10- This can be written as, {F1, F1, F1}, {F2, F2, F2},..., {F10, F10, F10}.
Another possibility is: All G-rings first, from F1-F10, then all S from F1-F10, then all B from F1-F10 -> {F1, F2,.., F10}, {F1, F2,..., F10}, {F1, F2,.., F10}.
And so on...
So all we need to calculate is all possible arrangements of 3 F1's, 3 F2's,... and 3 F10's (Total 3x10 = 30 objects).
Hence, \(\frac{30!}{(3!)^{10}} = \frac{30!}{6^{10}}\)