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by tarunajwani » Mon Nov 07, 2011 8:02 pm
A certain jewelry store sells customized rings in which three gemstones selected by the customer are set in the band of the ring. If exactly 5 different types of gemstone are available, and if at least two of the gemstones in any given ring must be different, how many different rings are possible?

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by rijul007 » Mon Nov 07, 2011 8:35 pm
5 different types of gemstone are available,
Customer needs to select 3 gem stones,
and at least two of the gemstones in any given ring must be different
this means at most 2 can be the same

__ __ __
All diff = 5*4*3 = 60
2 same = 5*4*1 * 3 = 60
Total no of arrangements = 120

120


Now because a ring has no front or back
A B C
is the same as
C B A

We divide it by 2
60

Is this the ans..??



Correction made
Last edited by rijul007 on Mon Nov 07, 2011 8:57 pm, edited 1 time in total.
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by shankar.ashwin » Mon Nov 07, 2011 8:47 pm
P(Atleast 2 colors) = 1 - P(same color)

Total possibility = 5*5*5 = 125.

Same rings= 5 ( 1 of each kind)

Therefore, p(Atleast 2 colors) = 125 - 5 = 120 IMO
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by user123321 » Mon Nov 07, 2011 8:58 pm
all are different 5p3 = 60
if two are different then 5c2*(3!/2!)*2 = 60
total 120..is that OA?

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by tarunajwani » Tue Nov 08, 2011 6:37 pm
Hey thanks guys !! Really appreciate your help !! 120 btw is the correct answer. Had a few silly questions in mind - if you could please help me with them - that would be great -

@ Rijul 007 - Earlier I did try solving the sum through the method you have used (the slot method) - understood the first part i.e. 5*4*3 - 60 and reached till the second part i.e. 5*4*1 ( but i do not understand why did you multiply by 3)-

@Shankar.aswhin & user 123321 - thanks guys !! Appreciate your help -
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by gmattaker2408 » Wed Nov 09, 2011 6:41 am
user123321 wrote:all are different 5p3 = 60
if two are different then 5c2*(3!/2!)*2 = 60
total 120..is that OA?

@user123231 where did you study your probability formula I saw you answered these probabilities questions well I was wondering if what books did you use to study? Thanks
Thanks!
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by user123321 » Wed Nov 09, 2011 7:09 am
gmattaker2408 wrote:
user123321 wrote:all are different 5p3 = 60
if two are different then 5c2*(3!/2!)*2 = 60
total 120..is that OA?

@user123231 where did you study your probability formula I saw you answered these probabilities questions well I was wondering if what books did you use to study? Thanks
I had probability in my academics. I forgot most of the stuff after I came out of the college :(. When I started my prep last month, I had gone through OG & Manhattan gmat books for basics. Did some problems from old threads in BTG. They helped a lot in understanding variety of problems. I read expert replies quite often. Thanks to them that helped in understanding tough questions.

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Want to do it right the first time.
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