Mixing
Combinatorics, distributing balls in boxes
$begingroup$
We've got the following problem:
Given $10$ boxes and $30$ balls such that $10$ balls are red, $10$ are blue and $10$ are green (balls of the same color are indistinguishable), how many ways are there to put the $30$ balls in the $10$ boxes knowing that some boxes may remain empty?
My attempt: First, I noticed that the problem doesn't specify if the boxes are distinguishable or not so I will try to answer for both cases, consider first the case when the boxes are distinguishable, I tried to break down the problem into 3 distribution problems, I did as follows:
We choose one of the three colors, say red, we have ${19 choose 9}$ ways to put the $10$ balls in the $10$ boxes.
The same is true no matter what color we choose, now, if I take one configuration of the red balls, I have ${19 choose 9}$ ways to add the blue balls to the "red filled" boxes, thus, to place red and blue in the boxes I would have: ${19 choose 9}^2$
By similar arguement when I add the green balls I would end up with ${19 choose 9}^3$ possibilities.
Now the second case, if the boxes are indistinguishable every time we count how to distribute a certain color we overcount by a factor of $10!$ so to fix that we should have $frac{{19 choose 9}^3}{10!^3}$
I may have done some silly mistake but I'm not sure since my textbook doesn't provide the solution to this problem.
EDIT: As pointed out in the comments, the indistinguishable case is not correct since we're not actually overcounting by 10!
combinatorics
asked Jan 31 at 18:32
Spasoje DurovicSpasoje Durovic
41811
$endgroup$
|
show 1 more comment
$begingroup$
We've got the following problem:
Given $10$ boxes and $30$ balls such that $10$ balls are red, $10$ are blue and $10$ are green (balls of the same color are indistinguishable), how many ways are there to put the $30$ balls in the $10$ boxes knowing that some boxes may remain empty?
My attempt: First, I noticed that the problem doesn't specify if the boxes are distinguishable or not so I will try to answer for both cases, consider first the case when the boxes are distinguishable, I tried to break down the problem into 3 distribution problems, I did as follows:
We choose one of the three colors, say red, we have ${19 choose 9}$ ways to put the $10$ balls in the $10$ boxes.
The same is true no matter what color we choose, now, if I take one configuration of the red balls, I have ${19 choose 9}$ ways to add the blue balls to the "red filled" boxes, thus, to place red and blue in the boxes I would have: ${19 choose 9}^2$
By similar arguement when I add the green balls I would end up with ${19 choose 9}^3$ possibilities.
Now the second case, if the boxes are indistinguishable every time we count how to distribute a certain color we overcount by a factor of $10!$ so to fix that we should have $frac{{19 choose 9}^3}{10!^3}$
I may have done some silly mistake but I'm not sure since my textbook doesn't provide the solution to this problem.
EDIT: As pointed out in the comments, the indistinguishable case is not correct since we're not actually overcounting by 10!
combinatorics
asked Jan 31 at 18:32
Spasoje DurovicSpasoje Durovic
41811
$endgroup$
$begingroup$
Seems fine to me, at least the distinguishable boxes case is definitely correct.
$endgroup$
– Shirish Kulhari
Jan 31 at 18:43
1
$begingroup$
The indistinguishable case is not correct, even for a single color. The case where you put one red ball in each of ten urns is not equivalent to any other cases, for instance.
$endgroup$
– lulu
Jan 31 at 18:45
$begingroup$
@lulu oh! that's right, I messed up, any idea how I could fix that?
$endgroup$
– Spasoje Durovic
Jan 31 at 19:07
$begingroup$
It's messy. Already for one color you have a partition function, see e.g. this (that one assumes every box gets a ball. If you want to drop that, then you get a sum of partition functions).
$endgroup$
– lulu
Jan 31 at 19:11
$begingroup$
@lulu I mean, this exercise was on the "easy" section of the book so I assume that the problem was about distinguishable boxes, but I'm gonna give the harder case a go nonetheless
$endgroup$
– Spasoje Durovic
Jan 31 at 19:16
|
show 1 more comment
$begingroup$
We've got the following problem:
Given $10$ boxes and $30$ balls such that $10$ balls are red, $10$ are blue and $10$ are green (balls of the same color are indistinguishable), how many ways are there to put the $30$ balls in the $10$ boxes knowing that some boxes may remain empty?
My attempt: First, I noticed that the problem doesn't specify if the boxes are distinguishable or not so I will try to answer for both cases, consider first the case when the boxes are distinguishable, I tried to break down the problem into 3 distribution problems, I did as follows:
We choose one of the three colors, say red, we have ${19 choose 9}$ ways to put the $10$ balls in the $10$ boxes.
The same is true no matter what color we choose, now, if I take one configuration of the red balls, I have ${19 choose 9}$ ways to add the blue balls to the "red filled" boxes, thus, to place red and blue in the boxes I would have: ${19 choose 9}^2$
By similar arguement when I add the green balls I would end up with ${19 choose 9}^3$ possibilities.
Now the second case, if the boxes are indistinguishable every time we count how to distribute a certain color we overcount by a factor of $10!$ so to fix that we should have $frac{{19 choose 9}^3}{10!^3}$
I may have done some silly mistake but I'm not sure since my textbook doesn't provide the solution to this problem.
EDIT: As pointed out in the comments, the indistinguishable case is not correct since we're not actually overcounting by 10!
combinatorics
asked Jan 31 at 18:32
Spasoje DurovicSpasoje Durovic
41811
$endgroup$
We've got the following problem:
Given $10$ boxes and $30$ balls such that $10$ balls are red, $10$ are blue and $10$ are green (balls of the same color are indistinguishable), how many ways are there to put the $30$ balls in the $10$ boxes knowing that some boxes may remain empty?
My attempt: First, I noticed that the problem doesn't specify if the boxes are distinguishable or not so I will try to answer for both cases, consider first the case when the boxes are distinguishable, I tried to break down the problem into 3 distribution problems, I did as follows:
We choose one of the three colors, say red, we have ${19 choose 9}$ ways to put the $10$ balls in the $10$ boxes.
The same is true no matter what color we choose, now, if I take one configuration of the red balls, I have ${19 choose 9}$ ways to add the blue balls to the "red filled" boxes, thus, to place red and blue in the boxes I would have: ${19 choose 9}^2$
By similar arguement when I add the green balls I would end up with ${19 choose 9}^3$ possibilities.
Now the second case, if the boxes are indistinguishable every time we count how to distribute a certain color we overcount by a factor of $10!$ so to fix that we should have $frac{{19 choose 9}^3}{10!^3}$
I may have done some silly mistake but I'm not sure since my textbook doesn't provide the solution to this problem.
EDIT: As pointed out in the comments, the indistinguishable case is not correct since we're not actually overcounting by 10!
combinatorics
combinatorics
asked Jan 31 at 18:32
Spasoje DurovicSpasoje Durovic
41811
asked Jan 31 at 18:32
Spasoje DurovicSpasoje Durovic
41811
edited Jan 31 at 19:10
Spasoje Durovic
asked Jan 31 at 18:32
Spasoje DurovicSpasoje Durovic
41811
asked Jan 31 at 18:32
Spasoje DurovicSpasoje Durovic
41811
asked Jan 31 at 18:32
Spasoje DurovicSpasoje Durovic
41811
41811
$begingroup$
Seems fine to me, at least the distinguishable boxes case is definitely correct.
$endgroup$
– Shirish Kulhari
Jan 31 at 18:43
1
$begingroup$
The indistinguishable case is not correct, even for a single color. The case where you put one red ball in each of ten urns is not equivalent to any other cases, for instance.
$endgroup$
– lulu
Jan 31 at 18:45
$begingroup$
@lulu oh! that's right, I messed up, any idea how I could fix that?
$endgroup$
– Spasoje Durovic
Jan 31 at 19:07
$begingroup$
It's messy. Already for one color you have a partition function, see e.g. this (that one assumes every box gets a ball. If you want to drop that, then you get a sum of partition functions).
$endgroup$
– lulu
Jan 31 at 19:11
$begingroup$
@lulu I mean, this exercise was on the "easy" section of the book so I assume that the problem was about distinguishable boxes, but I'm gonna give the harder case a go nonetheless
$endgroup$
– Spasoje Durovic
Jan 31 at 19:16
|
show 1 more comment
$begingroup$
Seems fine to me, at least the distinguishable boxes case is definitely correct.
$endgroup$
– Shirish Kulhari
Jan 31 at 18:43
1
$begingroup$
The indistinguishable case is not correct, even for a single color. The case where you put one red ball in each of ten urns is not equivalent to any other cases, for instance.
$endgroup$
– lulu
Jan 31 at 18:45
$begingroup$
@lulu oh! that's right, I messed up, any idea how I could fix that?
$endgroup$
– Spasoje Durovic
Jan 31 at 19:07
$begingroup$
It's messy. Already for one color you have a partition function, see e.g. this (that one assumes every box gets a ball. If you want to drop that, then you get a sum of partition functions).
$endgroup$
– lulu
Jan 31 at 19:11
$begingroup$
@lulu I mean, this exercise was on the "easy" section of the book so I assume that the problem was about distinguishable boxes, but I'm gonna give the harder case a go nonetheless
$endgroup$
– Spasoje Durovic
Jan 31 at 19:16
$begingroup$
Seems fine to me, at least the distinguishable boxes case is definitely correct.
$endgroup$
– Shirish Kulhari
Jan 31 at 18:43
$begingroup$
Seems fine to me, at least the distinguishable boxes case is definitely correct.
$endgroup$
– Shirish Kulhari
Jan 31 at 18:43
1
1
$begingroup$
The indistinguishable case is not correct, even for a single color. The case where you put one red ball in each of ten urns is not equivalent to any other cases, for instance.
$endgroup$
– lulu
Jan 31 at 18:45
$begingroup$
The indistinguishable case is not correct, even for a single color. The case where you put one red ball in each of ten urns is not equivalent to any other cases, for instance.
$endgroup$
– lulu
Jan 31 at 18:45
$begingroup$
@lulu oh! that's right, I messed up, any idea how I could fix that?
$endgroup$
– Spasoje Durovic
Jan 31 at 19:07
$begingroup$
@lulu oh! that's right, I messed up, any idea how I could fix that?
$endgroup$
– Spasoje Durovic
Jan 31 at 19:07
$begingroup$
It's messy. Already for one color you have a partition function, see e.g. this (that one assumes every box gets a ball. If you want to drop that, then you get a sum of partition functions).
$endgroup$
– lulu
Jan 31 at 19:11
$begingroup$
It's messy. Already for one color you have a partition function, see e.g. this (that one assumes every box gets a ball. If you want to drop that, then you get a sum of partition functions).
$endgroup$
– lulu
Jan 31 at 19:11
$begingroup$
@lulu I mean, this exercise was on the "easy" section of the book so I assume that the problem was about distinguishable boxes, but I'm gonna give the harder case a go nonetheless
$endgroup$
– Spasoje Durovic
Jan 31 at 19:16
$begingroup$
@lulu I mean, this exercise was on the "easy" section of the book so I assume that the problem was about distinguishable boxes, but I'm gonna give the harder case a go nonetheless
$endgroup$
– Spasoje Durovic
Jan 31 at 19:16
|
show 1 more comment
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$begingroup$
Seems fine to me, at least the distinguishable boxes case is definitely correct.
$endgroup$
– Shirish Kulhari
Jan 31 at 18:43
1
$begingroup$
The indistinguishable case is not correct, even for a single color. The case where you put one red ball in each of ten urns is not equivalent to any other cases, for instance.
$endgroup$
– lulu
Jan 31 at 18:45
$begingroup$
@lulu oh! that's right, I messed up, any idea how I could fix that?
$endgroup$
– Spasoje Durovic
Jan 31 at 19:07
$begingroup$
It's messy. Already for one color you have a partition function, see e.g. this (that one assumes every box gets a ball. If you want to drop that, then you get a sum of partition functions).
$endgroup$
– lulu
Jan 31 at 19:11
$begingroup$
@lulu I mean, this exercise was on the "easy" section of the book so I assume that the problem was about distinguishable boxes, but I'm gonna give the harder case a go nonetheless
$endgroup$
– Spasoje Durovic
Jan 31 at 19:16