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Suppose we have six marbles: 3 blue marbles, 2 red marbles,

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by swerve » Tue Dec 31, 2019 2:11 pm
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

The OA is B

Source: Magoosh
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by Scott@TargetTestPrep » Sat Jan 04, 2020 7:22 pm
swerve 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

The OA is B

Source: Magoosh
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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by deloitte247 » Sat Jan 04, 2020 9:51 pm
Blue marbles = 3
Red marbles = 2
Green marbles = 1
$$Total\ possible\ combinations=\frac{\left(no.\ of\ marbles\right)\left(no.\ of\ colors\right)}{\Pr oduct\ of\ no.\ of\ colors}$$
$$=\frac{6!\cdot3}{3!\cdot2!\cdot1!}$$
$$=\frac{6\cdot5\cdot4\cdot3!\cdot3}{3!\cdot2!\cdot1!}=\frac{6\cdot5\cdot4\cdot3}{2\cdot1}$$
$$=\frac{360}{2}=180$$
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marbles

by GMATGuruNY » Sun Jan 05, 2020 4:31 am
swerve 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

The OA is B

Source: Magoosh
Green marble:
Number of options = 3. (The black cup, the white cup, or the purple cup.)

Red marbles:
Case 1: 1 cup holds both red marbles
Number of options = 3. (The black cup, the white cup, or the purple cup.)
Case 2: 2 cups, each with 1 red marble
From 3 cups, the number of ways to choose 2 cups = 3C2 = (3*2)/(2*1) = 3
Total ways = Case 1 + Case 2 = 3+3 = 6

Blue Marbles:
Case 1: 1 cup holds all 3 blue marbles
Number of options = 3. (The black cup, the white cup, or the purple cup.)
Case 2: 1 cup with 2 blue marbles. another with 1 blue marble
From 3 cups, the number of ways to choose 1 cup to hold 2 blue marbles = 3C1 = 3
From the 2 remaining cups. the number of ways to choose 1 cup to hold the third blue marble = 2C1 = 2
To combine these options, we multiply:
3*2 = 6
Case 3: Each cup holds 1 blue marble
Number of options = 1. (1 blue marble in each cup.)
Total ways = Case 1 + Case 2 + Case 3 = 3+6+1 = 10

To combine our options for each color, we multiply:
3*6*10 = 180

The correct answer is B.
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