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mikemcgarry
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Suppose we have six marbles: 3 blue marbles, 2 red marbles, and one gr 02 Feb 2015, 17:44
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

95% (hard)

Question Stats:

29% (02:55) correct 71% (02:58) wrong based on 164 sessions

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Suppose we have six marbles: 3 blue marbles, 2 red marbles, and one green marble. Suppose we are going to put them into three cups: a black cup, a white cup, and a purple cup. The only restriction is that the two red marbles can’t be in the same cup. We could put as many as five (all except one of the reds) in any cup. We could leave one cup empty, or put some in each of the three cups. All combinations are allowed that don’t involve the two red marbles in the same cup. How many combinations are possible?

(A) 90
(B) 180
(C) 360
(D) 540
(E) 720

Show Spoiler

The GMAT loves tricky counting problems such as this. For a discussion of how to approach such problems, with the OE to this particular problem, see:
https://magoosh.com/gmat/2015/counting- ... -the-gmat/

Mike :-)
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A
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aviram
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Suppose we have six marbles: 3 blue marbles, 2 red marbles, and one gr 05 Feb 2015, 14:48
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mikemcgarry
Suppose we have six marbles: 3 blue marbles, 2 red marbles, and one green marble. Suppose we are going to put them into three cups: a black cup, a white cup, and a purple cup. The only restriction is that the two red marbles can’t be in the same cup. We could put as many as five (all except one of the reds) in any cup. We could leave one cup empty, or put some in each of the three cups. All combinations are allowed that don’t involve the two red marbles in the same cup. How many combinations are possible?
(A) 90
(B) 180
(C) 360
(D) 540
(E) 720

The GMAT loves tricky counting problems such as this. For a discussion of how to approach such problems, with the OE to this particular problem, see:
https://magoosh.com/gmat/2015/counting- ... -the-gmat/

Mike :-)
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Hey Mike,

The OA here is different from the one posted on the Magoosh link. I got 90, which is the OA here, but the site says 180.
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mikemcgarry
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Suppose we have six marbles: 3 blue marbles, 2 red marbles, and one gr 05 Feb 2015, 14:57
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aviram
mikemcgarry
Suppose we have six marbles: 3 blue marbles, 2 red marbles, and one green marble. Suppose we are going to put them into three cups: a black cup, a white cup, and a purple cup. The only restriction is that the two red marbles can’t be in the same cup. We could put as many as five (all except one of the reds) in any cup. We could leave one cup empty, or put some in each of the three cups. All combinations are allowed that don’t involve the two red marbles in the same cup. How many combinations are possible?
(A) 90
(B) 180
(C) 360
(D) 540
(E) 720

The GMAT loves tricky counting problems such as this. For a discussion of how to approach such problems, with the OE to this particular problem, see:
https://magoosh.com/gmat/2015/counting- ... -the-gmat/

Mike :-)

Hey Mike,

The OA here is different from the one posted on the Magoosh link. I got 90, which is the OA here, but the site says 180.
Show more
Dear aviram,
My friend, be careful. On the Magoosh site, there are two questions, #8 and #9, that superficially look identical, but there's a subtle different makes the answers different. The question posted here in this GC thread is #9, which indeed has the OA you calculated, but when you looked at the Magoosh website, I believe you looked at the OA of #8.
You never can read too carefully on the GMAT Quant section. Does all this make sense?
Mike :-)
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Suppose we have six marbles: 3 blue marbles, 2 red marbles, and one gr 19 Mar 2015, 13:30
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I tried this:

3 Blue Marble:

Different distribution : (1) All 3 in the same cup (2) Split 2-1 in two cups (3) Split 1-1-1

Case (1) 3 possibilities
Case (2) 6 possibilities ,
Case (3) 1 possibility

tot. 10 possibilities.

One Green Marble :

3 possibilities (one for each cup at the time)

First Red marble:

3 possibilities (one for each cup at the time)

Second Red marble:

2 possibilities ( because can't go in the same cup with the other Red marble)

TOTAL:

10*3*3*2 = 180 , answer, B .

Let me know if my conclusion is correct. thanks!
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Suppose we have six marbles: 3 blue marbles, 2 red marbles, and one gr 20 Mar 2015, 03:57
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mikemcgarry
Suppose we have six marbles: 3 blue marbles, 2 red marbles, and one green marble. Suppose we are going to put them into three cups: a black cup, a white cup, and a purple cup. The only restriction is that the two red marbles can’t be in the same cup. We could put as many as five (all except one of the reds) in any cup. We could leave one cup empty, or put some in each of the three cups. All combinations are allowed that don’t involve the two red marbles in the same cup. How many combinations are possible?
(A) 90
(B) 180
(C) 360
(D) 540
(E) 720

The GMAT loves tricky counting problems such as this. For a discussion of how to approach such problems, with the OE to this particular problem, see:
https://magoosh.com/gmat/2015/counting- ... -the-gmat/

Mike :-)
Show more


blue --> 3,0,0 ==> 3!/2! ; 2,1,0 ==> 3! ; 1,1,1 ==> 3!/3!
total = (3+6+1) = 10

red --> 1,1,0 ==> 3!/2! = 3

green --> 1,0,0 = 3!/2! = 3

answer = 10*3*3 = 90.
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Yosemite98
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Suppose we have six marbles: 3 blue marbles, 2 red marbles, and one gr 28 Jun 2025, 02:12
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mikemcgarry
Suppose we have six marbles: 3 blue marbles, 2 red marbles, and one green marble. Suppose we are going to put them into three cups: a black cup, a white cup, and a purple cup. The only restriction is that the two red marbles can’t be in the same cup. We could put as many as five (all except one of the reds) in any cup. We could leave one cup empty, or put some in each of the three cups. All combinations are allowed that don’t involve the two red marbles in the same cup. How many combinations are possible?

(A) 90
(B) 180
(C) 360
(D) 540
(E) 720

Show Spoiler

The GMAT loves tricky counting problems such as this. For a discussion of how to approach such problems, with the OE to this particular problem, see:
https://magoosh.com/gmat/2015/counting- ... -the-gmat/

Mike :-)
Show more
We have 3B, 2R, 1G.

1. How many ways can 3 identical blue marbels be distributed in 3 boxes? -> The easiest way is to do this is to assume that the buckets are also elements , thus we have to count the combinations of: B | B | B

C = 5!/(3!*2!) = 10

2. Red Marbels. We have to choose 2 boxes out of three for the marbels to be in -> 3!/2! -> 3

3. Green Marbel -> 3 options

total = 10*3*3 = 90 -> IMO A
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andreabylykbashi
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Suppose we have six marbles: 3 blue marbles, 2 red marbles, and one gr 30 Jun 2025, 03:23
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mikemcgarry
Suppose we have six marbles: 3 blue marbles, 2 red marbles, and one green marble. Suppose we are going to put them into three cups: a black cup, a white cup, and a purple cup. The only restriction is that the two red marbles can’t be in the same cup. We could put as many as five (all except one of the reds) in any cup. We could leave one cup empty, or put some in each of the three cups. All combinations are allowed that don’t involve the two red marbles in the same cup. How many combinations are possible?

(A) 90
(B) 180
(C) 360
(D) 540
(E) 720
Show more
We may proceed as follows.

Let (n,k) mean n choose k. The assignment of each set of colored balls is independent so we may find their individual numbers of distribution and multiply at the end. For each colored ball, we shall invoke the Bose-Einstein combinations (stars and bars), where we aim to put n indistinguishable objects (n=3,2,1) into k distinguishable boxes (k=3). For notation sake let [n,k] = (n+k-1,n), which is the Bose-Einstein number.

For the green ball, there's [1,3] = (1+3-1,1) = 3

For the red balls, there's [2,3] = (2+3-1,2) = (3,2) = (3,1) = 3

For the blue balls, we have [3,3] = (3+3-1,3) = (5,3) = (5,2) = 10

Hence there's 10*3*3 = 90 ways to arrange the balls. The answer is A.
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