In a bag are 6 identical green marbles, 4 identical blue marbles, and

Question

In a bag are 6 identical green marbles, 4 identical blue marbles, and 2 identical red marbles. If 3 marbles are picked at random from the twelve in the bag, how many distinct sets of three can be selected? 
a) 9
b) 10
c) 15
D) 21
E) 91

chetan2u · Accepted Answer

The sets can be distinct in following manner..

(1) ONE color of each - only one way -BGR

(2) TWO color of one and ONE of the any of the two remaining - So we can groups of 2 colors in 3C2 or 3 ways, but each way can have two combinations, for example GGR or RRG. 
So 3*2=6 ways - GGR, RRG, BBG, GGB, RRB, BBR

(3) All THREE of same colour - Red not possible as we have only 2 of them. so 2 ways - BBB, GGG

Total 1+6+2=9

A

ScottTargetTestPrep · Answer

There are so few ways to make the selections that it is easiest to simply list all the possibilities:

All 3 different colors: R, G, B - 1 way

2 different colors: RRG, RRB, BBR, BBG, GGR, GGB - 6 ways

All 3 the same color: BBB, GGG - 2 ways (note that we can’t have RRR because there are only 2 red marbles)

Thus, the total number of ways is 1 + 6 + 2 = 9.

Answer: A

GMATBusters · Answer

We need to pick three balls:

Let us put the picked 3 balls in 3 boxes as per color:
G, B, R

Now total no of picked balls is 3 is to be divided into 3 boxes: it can be seen as arranging 3 X (XXX) and 2 separators (|) in a line:
XX|X|X means there are 2 green, 2blue, and 1 red ball.

total number of arrangement = 5!/(3!2!) = 10

||XXX is the invalid case as we cannot have a selection of 3 red balls because only 2 red balls are present.

So, total valid selection = 10-1 = 9.

Answer A

(Question-based on  Similar logic  :  https://gmatclub.com/forum/how-many-positive-integers-less-than-10-000-are-there-in-85291.html  )

shameekv1989 · Answer

GMATBusters  :- Could you explain a bit more on the use of separator "| " (highlighted in red above) - When and how do we use these in questions?

GMATBusters · Answer

Hi

Bunuel has explained the approach beautifully in this question:  https://gmatclub.com/forum/how-many-positive-integers-less-than-10-000-are-there-in-85291.html .

If you have any doubt, feel free to tag me again.

AnirudhaS · Answer

I am just writing this answer to better understand myself. Basically this is an extension of  GMATBusters  's solution.

So let us assume we have 3 Green, 3 Blue and 3 Red balls (no restrictions to pick any colours). The main idea is as long as we have 3 balls of each there is no restriction, which means 6 green here just tells you you have "no restriction" on green, so the permutation GGG is allowed.

The other thing we have to realise is we are looking for distinct sets. This means GGR and GRG is the same, because it has 2 greens and 1 red. This is important.

Now Bunuel used "d" so I am going to use the same in his theory.

d = occurrence of a ball

We have 2 separators || and 3 buckets so as to speak _ _ _

Looks like this 
_|_|_

Now each bucket represents occurrence of Green, Blue or Red. (this is the part which is slightly different from Bunuel's problem)

Filling "d" in the first bucket means occurence of a green ball. Filling the second bucket with "d" means occurrence of a blue ball.
Therefore, ddd|| signifies the GGG occurrence.

I will list all possible occurrences, for those finding it hard to visualise
ddd|| --> 3G
|ddd| --> 3B
||ddd --> 3R
dd|d| --> 2G 1B
dd||d --> 2G 1R
d|dd| --> 1G 2B
|dd|d| --> 2B 1R
d||dd| --> 1G 2R
|d|dd| --> 1B 2R
d|d|d --> 1G 1B 1R

Which means we need to find out the permutation of (3+2) = 5 in a straight line. But the 2 separators and 3 occurrences are indistinguishable.

Therefore, required permutation =  \frac{5!}{3! 2!}  = 10

Now this is if we have no restrictions, remember we said 3G, 3B nd 3R
But in essence we only have 2 reds.

So the occurrence |||ddd will not happen. This is the RRR occurrence.

I hope this helps.

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In a bag are 6 identical green marbles, 4 identical blue marbles, and

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