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Tom is arranging his marble collection in a collectors case. [#permalink]

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10 Oct 2010, 03:08

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Difficulty:

95% (hard)

Question Stats:

46% (02:59) correct
54% (02:38) wrong based on 95 sessions

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Tom is arranging his marble collection in a collectors case. He has five identical cat-eyes, five identical sulphides, and three identical agates. If he can fit exactly five marbles into the case and must have at least one of each type, how many different ways can he arrange the case?

Tom is arranging his marble collection in a collectors case. He has five identical cat-eyes, five identical sulphides, and three identical agates. If he can fit exactly five marbles into the case and must have at least one of each type, how many different ways can he arrange the case?

(A) 120 (B) 150 (C) 420 (D) 1,260 (E) 1,680

I dont agree with the OA. Lets discuss.

Given the constraints, there is 2 ways to choose the stones :

1) Take 2 each of 2 types and 1 of the third type 3 ways to choose the type for which 1 stone is to be taken Ways to arrange = 5!/(2!x2!) = 30 Total = 30 x 3 = 90

2) Take 3 of 1 type, one each of the other types 3 ways to choose the type for which 3 stones are to be taken Ways to arrange = 5!/3! = 20 Total = 20 x 3 = 60

Damn... silly mistakes again. Thank you Shrouded. Feel embarrassed for making a mistake and posting it out thinking I was right. I feel I should delete the post.

Tom is arranging his marble collection in a collectors case. He has five identical cat-eyes, five identical sulphides, and three identical agates. If he can fit exactly five marbles into the case and must have at least one of each type, how many different ways can he arrange the case?

(A) 120 (B) 150 (C) 420 (D) 1,260 (E) 1,680

Don't forget KUDOS if you like this.

We have that 3 marbles CSA must be in the collection case as he must have at least one of each type.

Then five chosen marbles could be:

1. CSA-XX - meaning that other 2 marbles are the same (for example CSA-CC): \(\frac{5!}{3!}*C^1_3=60\) (\(C^1_3\) choosing which marble out of the 3 will be X and \(\frac{5!}{3!}\) arranging 5 marbles our of which 3 are identical).

2. CSA-XY - meaning that other 2 marbles are not the same (for example CSA-CS): \(\frac{5!}{2!2!}*C^2_3=60\) (\(C^2_3\) choosing which 2 marbles out of 3 will be X and Y and \(\frac{5!}{2!2!}\) arranging 5 marbles our of which 2 and other 2 are identical).

Re: Tom is arranging his marble collection in a collectors case. [#permalink]

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03 Dec 2013, 07:50

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Tom is arranging his marble collection in a collectors case. [#permalink]

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10 Dec 2013, 10:31

Either we could have 2,2,1 combo or 3,1,1 combo. Each of these combos in turn have 3 different cases. So, required answer is 3*(5!/3!) + 3*(5!/2!*2!) = 150.

Re: Tom is arranging his marble collection in a collectors case. [#permalink]

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02 Aug 2015, 20:24

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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