leve wrote:
Zeynep (the lady of the rings) has 10 identical gold, 10 identical silver and 10 identical bronze rings to wear in a particular ceremony. She will wear exactly one gold, one silver and one bronze (total of 3) rings on each of her 10 fingers such that for each finger she will wear one gold ring at the bottom, closest to the knuckle; then one silver ring; then one bronze ring in the most distal position. (According to the rituals of the ceremony this exact arrangement is required on each of her ten fingers.) 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?
(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.)
A) \(\frac{(10!)^3}{2^{10}}\)
B) \(\frac{(10!)^3}{3^{10}}\)
C) \(\frac{30!}{2^{10}}\)
D) \(\frac{30!}{3^{10}}\)
E) \(\frac{30!}{6^{10}}\)
Source: ITU GRE-GMAT prep seminar.
This question was presented as a possible very hard (top level) GRE-GMAT question. It's tricky but can be done under 2 minutes.
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 on2. 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 Bunuelcan something of this sort be expected on the GMAT?