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5 Balls in 3 boxes (Permutation Combination)

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5 Balls in 3 boxes (Permutation Combination)

by Stuti567 » Thu May 05, 2016 11:16 pm

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Five balls of different colours are to be placed in three boxes of different sizes. Each box can hold five balls. The number of ways in which we can place the balls in the boxes so that no box remains empty is?

a) 30
b) 150
c) 600
d) 900

OA b

[spoiler]I solved by first placing the balls in the three different boxes in 3 x 2 x 1 ways, then the other two balls in the 3 boxes in 3 x 3 ways, so I'm getting the answer as 54! [/spoiler]

Please explain why my method is wrong and what's the correct method.
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by GMATGuruNY » Thu May 05, 2016 11:34 pm
Stuti567 wrote:Five marbles of different colors are to be placed in three boxes of different sizes. Each box can hold five marbles. The number of ways in which we can place the marbles in the boxes so that no box remains empty is?

a) 30
b) 150
c) 600
d) 900
Case 1: 1 box has 3 marbles, the other 2 boxes each have 1 marble
Number of box options for the box with 3 marbles = 3. (Any of the 3 boxes.)
For this box, the number of ways to choose 3 marbles from 5 options = 5C3 = (5*4*3)/(3*2*1) = 10.
Number of marble options for the next box = 2. (Either of the 2 remaining marbles.)
Number of marble options for the last box = 1. (Only 1 marble left.)
To combine these options, we multiply:
3*10*2*1 = 60.

Case 2: 1 box has 1 marble, the other 2 boxes each have 2 marbles
Number of box options for the box with 1 marble = 3. (Any of the 3 boxes.)
Number of marbles that could be placed in this box = 5. (Any of the 5 marbles.)
From the 4 remaining marbles, the number of ways to choose 2 marbles for the next box = 4C2 = (4*3)/(2*1) = 6.
From the 2 remaining marbles, the number of ways to choose 2 marbles for the last box = 2C2 = (2*1)/(2*1) = 1.
To combine these options, we multiply:
3*5*6*1 = 90.

Total ways = 60+90 = 150.

The correct answer is B.
I solved by first placing the balls in the three different boxes in 3 x 2 x 1 ways
The product in red seems to imply that the first ball has 3 choices (any of the 3 boxes), while the second ball has 2 choices (either of the 2 remaining boxes), and the third ball has only 1 choice (the one remaining box).
This product overly constrains the problem, since it is possible that the first 3 balls are all placed in the SAME BOX or are placed in TWO DIFFERENT BOXES.
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by Stuti567 » Fri May 06, 2016 2:35 am
GMATGuruNY wrote:
Stuti567 wrote:Five marbles of different colors are to be placed in three boxes of different sizes. Each box can hold five marbles. The number of ways in which we can place the marbles in the boxes so that no box remains empty is?

a) 30
b) 150
c) 600
d) 900
Case 1: 1 box has 3 marbles, the other 2 boxes each have 1 marble
Number of box options for the box with 3 marbles = 3. (Any of the 3 boxes.)
For this box, the number of ways to choose 3 marbles from 5 options = 5C3 = (5*4*3)/(3*2*1) = 10.
Number of marble options for the next box = 2. (Either of the 2 remaining marbles.)
Number of marble options for the last box = 1. (Only 1 marble left.)
To combine these options, we multiply:
3*10*2*1 = 60.

Case 2: 1 box has 1 marble, the other 2 boxes each have 2 marbles
Number of box options for the box with 1 marble = 3. (Any of the 3 boxes.)
Number of marbles that could be placed in this box = 5. (Any of the 5 marbles.)
From the 4 remaining marbles, the number of ways to choose 2 marbles for the next box = 4C2 = (4*3)/(2*1) = 6.
From the 2 remaining marbles, the number of ways to choose 2 marbles for the last box = 2C2 = (2*1)/(2*1) = 1.
To combine these options, we multiply:
3*5*6*1 = 90.

Total ways = 60+90 = 150.

The correct answer is B.
I solved by first placing the balls in the three different boxes in 3 x 2 x 1 ways
The product in red seems to imply that the first ball has 3 choices (any of the 3 boxes), while the second ball has 2 choices (either of the 2 remaining boxes), and the third ball has only 1 choice (the one remaining box).
This product overly constrains the problem, since it is possible that the first 3 balls are all placed in the SAME BOX or are placed in TWO DIFFERENT BOXES.
Hi,

Thanks for your solution. It does help.

But I still do not get why my process is wrong and how would the question be different for this method. Is it because here both the boxes and the balls are different?

Also, why can't I first select 3 balls for the boxes by 5C3, then put the other 2 balls in any of the 3 boxes using 3 x 3 = 90.

I also used this other method by divided the question into stages:
Stage 1: Choose 3 balls and distribute into 3 boxes in 5C3 x 3! = 60 ways
Stage 2: Choose another ball and put in into one of the 3 boxes = 2C1 x 3 = 6 ways
Stage 3: Take the another ball and put it into any of the 3 boxes= 3 ways

Now , multiplying all we get 60 x 6 x 3

This seems wrong but I don't understand why!

What level GMAT question would it be?
Please elaborate :)
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by GMATGuruNY » Fri May 06, 2016 4:48 am
Stuti567 wrote:I still do not get why my process is wrong and how would the question be different for this method. Is it because here both the boxes and the balls are different?
Yes.
Because the balls and the boxes are all different, we must be careful not to overly constrain the number of options for each ball and box.

Let's say that the first two balls selected are as follows:
A, B.
I solved by first placing the balls in the three different boxes in 3 x 2 x 1 ways
This approach assumes the following:
Number of options for A = 3. (Any of the 3 boxes)
Number of options for B = 2. (Either of the 2 remaining boxes)
The assumption for B is invalid.
Since it is possible that A and B are placed in the same box, it is not true that the number of options for B = 2.
Also, why can't I first select 3 balls for the boxes by 5C3, then put the other 2 balls in any of the 3 boxes using 3 x 3 = 90
None of the 3 boxes may remain empty.
Thus:
Once 3 balls are placed in a box, the 4th ball MUST be placed in one of the 2 remaining empty boxes (yielding 2 options), with the result that the 5th ball MUST be placed in the one remaining empty box (yielding 1 option):
5C3 * 2 * 1 = 60.
I also used this other method by divided the question into stages:
Stage 1: Choose 3 balls and distribute into 3 boxes in 5C3 x 3! = 60 ways

This seems wrong but I don't understand why!
Let's say that the 3 balls selected are as follows:
A, B, C.
Here again, the product in red -- 3*2*1 -- assumes the following:
Number of options for A = 3. (Any of the 3 boxes)
Number of options for B = 2. (Either of the 2 remaining boxes)
Number of options for C = 1. (The one remaining box)
The values for B and C are invalid, since A, B and C could all be placed in the SAME box.
What level GMAT question would it be?
Please elaborate :)
700+.
More complex than a typical GMAT problem involving combinatorics.
I would consider this problem LOW PRIORITY.
Unless you've mastered all other aspects of the GMAT, your time is better spent elsewhere.
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by Stuti567 » Fri May 06, 2016 5:02 am
@GMATGuruNY
Thanks a lot! All of it makes perfect sense now :)
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by MartyMurray » Fri May 06, 2016 5:11 am
Stuti567 wrote:I still do not get why my process is wrong and how would the question be different for this method. Is it because here both the boxes and the balls are different?

Also, why can't I first select 3 balls for the boxes by 5C3, then put the other 2 balls in any of the 3 boxes using 3 x 3 = 90.

I also used this other method by divided the question into stages:
Stage 1: Choose 3 balls and distribute into 3 boxes in 5C3 x 3! = 60 ways
Stage 2: Choose another ball and put in into one of the 3 boxes = 2C1 x 3 = 6 ways
Stage 3: Take the another ball and put it into any of the 3 boxes= 3 ways

Now , multiplying all we get 60 x 6 x 3

This seems wrong but I don't understand why!
One issue with your first method is that the three balls that you distributed into the three boxes could actually have been put into one of the three boxes or distributed into two boxes so that there are two in one and one in another. So you ruled out multiple possible scenarios.

The foundation of the problem with that method is that, yes, the three balls and three boxes are all unique. So putting a particular ball into a particular box is not the same as putting a different ball in that box. So whatever method you use has to work with each ball individually so that any ball can end up in any box, possibly with some other ball.

Meanwhile, your methods create too many outcomes because you are choosing the balls repeatedly. First you choose three balls to put into the three boxes, and then you choose from the remaining two, but guess what, you are always putting all five into the boxes. Look at the second and third steps of your second method to easily see the issue. You choose one of two, and then choose a box to put it in. Then guess what, you choose the other one and put it in a box. So actually either way you choose BOTH them! So the 2C1 part does not make sense.

So your methods are partly based on the idea that not only does which ball goes into which box matter but also the order in which the balls are placed in the boxes matters, and actually the order in which the balls are placed in the boxes does not matter. So choosing which go in first, second and third does not work. The different outcomes have to be created by putting different balls into different boxes without being concerned with the order in which the balls are placed into the boxes.
What level GMAT question would it be?
This is a 700 level question.
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Re: 5 Balls in 3 boxes (Permutation Combination)

by Saumyojit » Fri Jul 02, 2021 1:26 pm
I have discussed here but i did not understand it .



Possible distribution sizes are (3,1,1) and (2,2,1)

Lets go with (3,1,1) .. suppose triplets chosen are b1 b2 b3 , singles are b4, b5.
SO with a particular combination of distribution sizes, there are 3! arrangements



My interpretation and approach is that :
Largest box second larger box Smallest box
B1 B2 B3 B4 B5
B1B2B3 B5 B4
B4 B1B2 B3 B5
B5 B1B2 B3 B4
B5 B4 B1 B2 B3
B4 B5 B1 B2 B3

Let me choose 3 balls out of 5 --> 5C3 , then choose 1 ball out of 2--> 2c1 , then choose
1 ball out of 1 ->1
Now they will arrange Among the Boxes in 3! ways

5c3 * 2c1 *1 *3! = 120
Similarly goes for (2,2,1) distribution groups .
Then, I will add both the Cases.


suppose Just take this two events,

when triplets is at largest box , one blue color ball is at 2nd larger box and one black ball at smallest box
Now that blue color ball can be at the Smallest box and black at 2nd larger keeping triplets at Largest box .

Both the above arrangement is unique right ? The location of triplets is same in the above two cases but the location of blue and black ball is changing.

Why this counted as twice?
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