In how many ways can 5 different colored marbles be placed in 3 distin

We have 5 marbles and 3 pockets
So we have two cases

Case-1: One Pocket with 3 marbles and two pockets with 1 marble each
No. of Arrangements = 5C3 * 3C1 * 2! = 10*3*2 = 60
5C3 - No. of ways of choosing 3 out of 5 marbles which have to go in one pocket
3C1 - No. of ways of choosing 1 out of 3 pockets in which 3 marbles have to go
2! - No. of ways of arranging remaining 2 marbles between remaining two pockets which get one marble each


Case-2: Two Pockets with 2 marbles each and one pockets with 1 marble
No. of Arrangements = 5C2 * 3C2 * 3C2 = 10*3*3 = 90
5C2 - No. of ways of choosing 2 out of 5 marbles which have to go in one pocket
3C2 - No. of ways of choosing 2 out of remaining 3 marbles which have to go in second pocket
3C2 - No. of ways of choosing 2 out of 3 pockets in each of which 2 marbles have to go

Total cases = 60+90 = 150

Answer: option D

GMATinsight, Bunuel, VeritasPrepKarishma,

I'm hoping one of you can explain why the 1-1-1 combination is ignored in answering this question. The question asks:
Would the answer not include 5P3? Since we can have at least 1 marble (i.e. exactly one marble) in each spot and the question does not specify that we must use all five of the marbles. Additionally, I think that permutation is the right way of counting for the 1-1-1 combination since a marble arrangement of Green-Blue-Red in pockets one-two-three is different from Blue-Red-Green in pockets one-two-three and so forth.
For the 1-1-1 combination, we will have: 5!/(5-3)! = 60 different arrangements.

Then we come to the 3-1-1 and 2-2-1 combinations.

For the 3-1-1 combination, in pocket one we can have any three of the five distinct marbles. The order inside one pocket doesn't matter. Therefore, we will use 5C3. For the 2nd pocket, we have two marbles left and we can only pick one because we must leave one for the third pocket. Therefore, we will use 2C1. And for the last (third) pocket we only have one choice. Now, we have 5C3, 2C1 and 1C1 in pockets one-two-three. These can be ordered in 3!/2! different ways because 2C1 = 1C1, and we have already counted the possibility of different colors in each pocket.
So for 3-1-1 combination, we have: 5C3 * 2C1 * 1C1 * 3!/2! = 60 different arrangements.

Similarly, for the 2-2-1 combination we have: 5C2 * 3C2 * 1C1 * 3!/2! = 90 different arrangements.

Finally, we will get 60+60+90 = 210 arrangements if we consider 1-1-1 to be a valid option - since the question does not explicitly exclude this possibility.

Could you please explain why we ignored 1-1-1 combination?

Hey jhabib,

You have to place all the marbles. If you assume the 1-1-1 combination, note that 2 marbles are leftover. But you HAVE TO distribute 5 marbles. So 3-1-1 and 2-2-1 are the only possibilities. All 5 marbles must be distributed.

Hi jhabib

The Question has clearly specified that "ALL the Marbles have to be assigned to 3 pockets" so 1-1-1 needs to be ignored

Along with 1-1-1, 1-1-2, 1-2-1 ans 2-1-1 also need to be ignored.

I hope it helps!

hi

I have seen a solution to a problem similar to this one elsewhere on the forum
let me explain it to you

5 different colored marbles can be placed in 3 distinct pockets without any restriction is

= 3^5 = 243

as we are asked to find out the ways in which any pocket must get at least 1 marble, lets find out the ways in which any pocket must not get at least 1 marble, and then subtract the number of ways in which any pocket must not get at least 1 marble from the total number of ways, that is 3^5

so lets get going

number of ways in which all marbles can get to 1 pocket is

= 3, as there are 3 distinct pocket in total

now, number of ways in which 2 pockets can get all the marbles and 1 pocket remains empty is

= (2^5 - 2) * 3 = 90

2 has been subtracted to eliminate the possibility that any pocket out of 2 can get all the marbles

3 has been multiplied with the whole expression, because 2 pockets out of 3 have been selected

now we are in business

the answer is

= 243 - 90 - 3

= 150 (D)

hope this helps
thanks

cheers, and do consider some kudos, man

Dear Karishma ..
I am bitterly confused with this problem . I doubt the approach or the problem has some defect . Please correct me if I am wrong .
All the approaches are considering 3 bags as identical , but in problem they are distinct .
My approach to this problem :
I am considering 5 different marbles a b c d e and 3 different bags 1 2 3 .
So first distribute 3 balls one ball each to each bag .
select 3 balls to distribut 5C3 and then distribute to 3 different bags in 3! ways .
so total 60 ways .
Now we are left with 2 balls and 3 bags each containing one ball each .
for first ball we have 3 options and second ball again 3 options . 3*3 ways .
TOTAL = 60*3*3=540 ways .

Please somebody help me in this . This problem just made me mad ..

You have some double counting here. When you select some from a group and distribute and then distribute the rest to the same bags/pockets, there is double counting.

Say you distributed a, b, c first such that
Bag1 had a
Bag2 had b
Bag3 had c
Then you distributed d and e such that Bag1 got d and bag2 got e.

Take another case.
Say you distributed d, b and c first such that
Bag1 had d
Bag2 had b
Bag3 had c
Then you distributed a and e such that Bag1 got a and bag2 got e.

Note that both cases have exactly the same end result. But you would count them as two separate cases. Similarly, there will be other cases which will be double counted. Hence this method will not work.

We have two cases: The first case has one pocket with 3 marbles, 1 marble in each of the two remaining pockets. The second case has two pockets with 2 marbles each, and one pocket with one marble. We always add cases together to obtain the total number of possibilities.

Case 1: (3,1,1) First choose which of the 3 pockets will get three marbles. There are 3C1 =3 ways to do this. Then, choose the number of ways we can place 3 marbles into that box from 5 marbles. There are 5C3 = 10 ways to do this. Next we must place a marble in the next to the last box. There are 2C1 = 2 ways to do this. Then there is 1C1 = 1 way to place the last marble into the last pocket. Each time we make a subsequent selection when working through a case we multiply, so there are 3*10*2*1= 60 ways to place the marbles in case 1

Case 2: (2,2,1) First we must select which two pockets get two marbles each. There are 3C2 = 3 ways to do this. Then, we must choose two marbles to go in the first of two pockets. There are 5C2=10 ways to do this. From the 3 remaining marbles, there are 3C2 = 3 ways to place two marbles in the second pocket. Finally there is 1C1 =1 way to place the last marble in the last box. Multiplying, we have 3*10*3*1 = 90 ways to place the marbles in case two.

Thus there are 60+90 = 150 ways we can place the marbles

Solved it a little differently than others from what I read . So will share.

I first made sure that each of the pocket has at least 1 marble.
That is 5C3 * 3!

Then you can either distribute the remaining 2 as 2, 0, 0 or 1, 1, 0
That is 3C1 or 3C2 * 2 ! (3C1 is the no. of ways of choosing a pocket to put everything in. )

That gives 150.
D !

Solution attached

Hi VeritasKarishma

Why do we multiply the selection by 3 and not 3!

in case 1 (3,1,1) no. of ways= 5C3*2C1*3! (5C3 to select 3 out of 5 diff combo of marbles, 2C1 to select 1 marble and 3! to arrange)

Pls explain!

Because the arrangement is (3, 1, 1) so all you need to do is pick one pocket which will get 3 marbles. Both others will get 1 marble each automatically.

So you pick 1 pocket in 3C1 ways.
You pick 3 of the 5 marbles in 5C3 ways and give them to that pocket.
Now you have 2 pockets and 2 marbles. You can put them 1 marble in 1 pockets in 2 ways: MarbleA goes into P2 and MarbleB goes into P3 or MarbleA goes into P3 and MarbleB goes into P2.

So 3 * 5C3 * 2

My only question here is that why are you not multiplying the second scenario i.e., 2-2-1 with another 2!?
So, you can select 2 out of 5 marbles in 5C2 ways.
Another 2 in 3C2 ways.
Last one in 1 way.
Now, instead of finding two places for the 2 and 2 marbles, why didn't you just multiply by 3! instead? To find the number of arrangements?

GMATinsight

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