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

MAGOOSH OFFICIAL SOLUTION:

First, consider possibilities the 3 blue marbles.

Blue Case I: all three in one cup = 3 possibilities

Blue Case II: two in one cup, one in another è three choices for the cup with two, then two choices for the cup with one = 6 possibilities

Blue Case III: one in each cup = 1 possibility

total = 3 + 6 + 1 = 10 possibilities for the blue marbles

Now, the 2 red marbles.

Red Case I: two in one cup = 3 possibilities

Red Case II: one in cup, one in another = 3 possibilities for which cup is empty

total = 6 possibilities for the red marbles

Now, the green marble can simply go in one of the three cups = 3 possibilities.

By the Fundamental Counting Principle, multiple the blue & red & green possibilities:

N = 10*6*3 = 180

Answer = (B)
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# ways to distribute Blue marbels(3) is :
1 in each cup= 1
2 in one cup and 1 in other=6
all 3 in one cup = 3
total 1+6+3=10 possibilities
# ways to distribute Red marbels(2) is :
1 in each cup = 3 ways
all in one cup = 3ways
total 6ways.

# ways to distribute Green marbels(1) is :
3 ways
total number of ways = 10*6 *3 = 180

Alternately this can be solved using combitronics with possible distribution
(6,0,0) one cup receives all = 3ways
(5,1,0)= one cup receives 5 and other receives 1 ; 3C1 * 2 but we will have to multiply this by 3! ; total = 3*2*6=36
(4,1,1)= one cup receives 4 and other receive 1 each ; 4C1 *2*1 but we will have to multiply this by 3!/2! ; total =24
like wise if we do for
4,2,0
3,3,0
3,2,1
2,2,2
we will get same total 180 ways.



Answer is B.

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.

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

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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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.