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Re: Suppose we have six marbles: 3 blue marbles, 2 red marbles, and one [#permalink]
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Bunuel wrote:
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. We could put all six in any cup and leave two cups empty; or we could put marbles in two cups and leave one cup empty; or we could put some marbles in each of the three cups. How many combinations are possible?

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


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I always get confused with such kind of problems, letters going to letterboxes , keys going to locks , gifts distributed to children , bulbs switching no.of lights on.
Can you explain basic concept in this ? or provide link where i can get this.
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Re: Suppose we have six marbles: 3 blue marbles, 2 red marbles, and one [#permalink]
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I think it would be easier to think in terms of 6 marbles fitting into 3 cups, then dividing for marble colors.

As such:
(6!*3)/3!*2! = 180
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Re: Suppose we have six marbles: 3 blue marbles, 2 red marbles, and one [#permalink]
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Bunuel wrote:
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. We could put all six in any cup and leave two cups empty; or we could put marbles in two cups and leave one cup empty; or we could put some marbles in each of the three cups. How many combinations are possible?

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


Kudos for a correct solution.


The green marble can be in the black, white or purple cup; thus there are 3 possibilities for the green marble.

The black cup can contain 0, 1 or 2 red marbles. If it contains 2 red marbles, then none of the remaining cups can contain any red marbles (one possibility). If it contains 1 red marble, then the remaining red marble can be either in the white or purple cup (two possibilities). If the black cup contains no red marbles, then the white cup can contain 0, 1 or 2 red marbles (and the purple cup will contain any remaining red marbles; thus three possibilities). The red marbles can be distributed in 1 + 2 + 3 = 6 ways.

Let’s do the same for the blue marbles. The black cup can contain 0, 1, 2 or 3 blue marbles. If it contains 3 blue marbles, then none of the remaining cups can contain any blue marbles (one possibility). If it contains 2 blue marbles, then the remaining blue marble can be either in the white or purple cup (two possibilities). If the black cup contains one blue marble, then the white cup can contain 0, 1 or 2 blue marbles (three possibilities). Finally, if the black cup contains no blue marbles, the white cup can contain 0, 1, 2 or 3 blue marbles (and the purple cup will contain any remaining blue marbles, four possibilities). The blue marbles can be distributed in 1 + 2 + 3 + 4 = 10 ways.

In total, there are 3 x 6 x 10 = 180 ways to distribute the marbles.

Answer: B
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Re: Suppose we have six marbles: 3 blue marbles, 2 red marbles, and one [#permalink]
No. of ways to distribute each set of identical colored marbles (using Star-Bar):
Blue: 5C2 = 10
Red : 4C2 = 6
Green : 3C2 = 3
Now by principle of counting, total ways = 10*6*3 = 180.
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Re: Suppose we have six marbles: 3 blue marbles, 2 red marbles, and one [#permalink]
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