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Ring - Gems

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by ronnie1985 » Tue May 22, 2012 10:02 am
A jewelry store sells customized rings in which 3 gems selected by the customer are set in a straight row along the band of the ring. If exactly 5 different gems are available and if at least 2 gems in any given ring must be different, how many different rings are possible?


20
60
90
120
210


Please explain, how to solve.
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by aneesh.kg » Tue May 22, 2012 11:05 am
ronnie1985 wrote:A jewelry store sells customized rings in which 3 gems selected by the customer are set in a straight row along the band of the ring. If exactly 5 different gems are available and if at least 2 gems in any given ring must be different, how many different rings are possible?


20
60
90
120
210
Hi ronnie1985,

Lets say that the gems available are A,B,C,D,E.

Total No. of Rings
= No. of rings will three different gems + No. of rings with 2 different gems
= (Arrange any 3 of the 5 gems) + (Select any of 2 of the 5 gems)*(Arrange them)
= 5P3 + (5C2)*(Select one of the two that you want to repeat)*(Arrange the three gems which have two identical gems)
[Say you select A and B then you might want to repeat A (in a AAB, ABA, etc fashion) or B(in a ABB, BAB, etc fashion]
= 5P3 + (5C2)*(2C1)*(3!/2!)
= 60 + 60
= 120

[spoiler](D)[/spoiler] is the correct answer.
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by GMATGuruNY » Wed May 23, 2012 2:42 am
ronnie1985 wrote:A jewelry store sells customized rings in which 3 gems selected by the customer are set in a straight row along the band of the ring. If exactly 5 different gems are available and if at least 2 gems in any given ring must be different, how many different rings are possible?


20
60
90
120
210
Please explain, how to solve.
Good rings = total possible rings - bad rings.

Total possible rings:
For each of the 3 positions in the ring, any of the 5 gems could be selected.
Number of options for the first position = 5.
Number of options for the second position = 5.
Number of options for the third position = 5.
To combine these options, we multiply:
5*5*5 = 125.

Bad rings:
In a good ring, at least 2 of the gems are different.
Thus, in a bad ring, all 3 gems are the same.
Thus, if the gem types are A, B, C, D and E, there are only 5 bad rings:
AAA, BBB, CCC, DDD, EEE.

Good rings = 125-5 = 120.

The correct answer is D.
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by ankita1709 » Sat May 26, 2012 9:23 am
What if I do it like this
Case 1- 2 different stones
5C2* ways of arranging(i.e. 3!/2!)
=30
Case 2- 3 different stones
5C3*ways of arranging(i.e. 3!)
=60

Hence total = 90
Please tell me what am I missing
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by aneesh.kg » Sat May 26, 2012 9:35 am
ankita1709 wrote:What if I do it like this
Case 1- 2 different stones
5C2* ways of arranging(i.e. 3!/2!)
=30
Case 2- 3 different stones
5C3*ways of arranging(i.e. 3!)
=60

Hence total = 90
Please tell me what am I missing
Ankita,

The problem is in Case 1.
After the 5C2, you also have to include a 2C1 to select the one gem out of the two gems that you would like to have twice in the ring. This normally escapes one's attention.

Say, out of A, B, C, D and E, you select A and B by 5C2 then you might want to repeat A twice (AAB, ABA,..) or B twice (BBA, BAB..). That's why 5C2 * 2C1 * (3!/2!) is the correction.

Please read my post above. I've solved this problem using the same method.

Having said that, the method shown by Mitch is the better solution.
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by ankita1709 » Sat May 26, 2012 9:40 am
Ohhhhh .. got it
Thanks a lot.:)
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