Combination / Permutation

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by analyst218 » Thu Apr 15, 2010 2:01 pm
pkw209 wrote:Image
hm.


for one ring where all of the stones are diff. there's 5c3=60
for 1 same 2 diff. there's.. 5x2x1= 10

don't get it.
please explain

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by pkw209 » Thu Apr 15, 2010 2:11 pm
I think the answer is 120.

Total number of possibilities for the 3 spots if all 5 gemstones were allowed = 5 x 5 x 5 = 125

However, 2 of the same or all 3 different gemstones are only allowed so you need to subtract any combination that has all 3 of the same gemstones.

125-5=120

Please correct me if I'm wrong.

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by analyst218 » Thu Apr 15, 2010 2:19 pm
pkw209 wrote:I think the answer is 120.

Total number of possibilities for the 3 spots if all 5 gemstones were allowed = 5 x 5 x 5 = 125

However, 2 of the same or all 3 different gemstones are only allowed so you need to subtract any combination that has all 3 of the same gemstones.

125-5=120

Please correct me if I'm wrong.
Well the question states exactly 5 different gemstones, which means 1 each thus not allowed to repeat..

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by pkw209 » Thu Apr 15, 2010 2:29 pm
"if at least two of the gemstones in any given ring must be different"

x - x - y still satisfies this. While x stone is repeated, two different stones are also on the ring.

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by analyst218 » Thu Apr 15, 2010 2:38 pm
pkw209 wrote:"if at least two of the gemstones in any given ring must be different"

x - x - y still satisfies this. While x stone is repeated, two different stones are also on the ring.
do u have the OA

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by pkw209 » Thu Apr 15, 2010 2:41 pm
Sorry, I pulled this from a random source although I believe I have seen it either on a Kaplan or MGMAT test.

I'm sure someone will verify the answer soon.

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by KapTeacherEli » Thu Apr 15, 2010 7:58 pm
This is a Kaplan one, and PKW nailed the solution.

Remember, when dealing with complex permutations, a common trap is to work super hard on breaking down all the little outcomes to solve bit by bit. We could figure out the number of combinations of 3 different gemstones, then figure out the number of organizations of 2 and 1 (xxy, xyx, yxx) and work from there as well. But it's much simpler to figure out what DOESN'T count: three of the same stone.

So, since there are 5 possibilities for the first stone, 5 for the second, and 5 for the third, that gives us 5 x 5 x 5 gemstone combinations. But, some don't count: those with no stones different. Clearly, there are five total invalid combinations, one for each of the five types of stone. so, 125 - 5 = 120 possible rings.
Eli Meyer
Kaplan GMAT Teacher
Cambridge, MA
www.kaptest.com/gmat

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by eaakbari » Thu Apr 15, 2010 10:38 pm
Well even though I like pkws solution, I am not sure if its correct, as the question has asked us how many rings.
If we have red blue and white denoted by rbw

The ring rbw is the same as wbr

The solution does not account for that
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by eaakbari » Thu Apr 15, 2010 10:41 pm
On second thought since the answer option 40 is not there, your solution must be correct. But I must say that the question had a degree of ambiguity in it.
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