# 3 cans to be handed out to three people

After you hand out the first can, there are no restrictions on handing out the second can. Three places to put the first can, and three places to put the second can. 3^2 possible choices. If the cans are distinguishable you are done. If the cans are indistinguishable you have to remove the double counting of the degenerate cases. There is only one way for anyone person to get both cans, so three cases are not double counted. The remaining 6 get reduced to three for a total of six.

## Homework Statement

Hi say I have 2 identical tin cans and I want to hand it out to three people B,C,D

I have three ways of doing this by giving all 2 cans to person B, then to person C, then to person D.

Then I have another 2 ways by giving person B 1 can, and C one can. Also person B 1 can, and person D one can.

So far I have a total of 5 ways.

The only way that is left is to give person D one can, and person C one can.

So I get a total of 6 ways.

## The Attempt at a Solution

I am trying to find a generalized way of doing this (not having to draw people and cans.)

It seems to me that this is different than a permutation, and combination.

For example handing out 3 cans to 3 people yields 10 possible ways.

Is there any topic that I can read that will introduce me to how I can do this?

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## Homework Statement

Hi say I have 2 tin cans and I want to hand it out to three people B,C,D

I have three ways of doing this by giving all 2 cans to person B, then to person C, then to person D.

Then I have another 2 ways by giving person B 1 can, and C one can. Also person B 1 can, and person D one can.

So far I have a total of 5 ways.

The only way that is left is to give person D one can, and person C one can.

So I get a total of 6 ways.

## The Attempt at a Solution

I am trying to find a generalized way of doing this (not having to draw people and cans.)

It seems to me that this is different than a permutation, and combination.

For example handing out 3 cans to 3 people yields 10 possible ways.

Is there any topic that I can read that will introduce me to how I can do this?

Are the cans identical, or are they distinguishable? For example, if the cans are different we can have two distinct arrangements where person A gets can #1 and person B gets can #2, versus person A getting can #2 and Person B getting can #1? If the cans are identical, there is just one such possibility: persons A and B each get one can.

Ray Vickson said:
Are the cans identical, or are they distinguishable? For example, if the cans are different we can have two distinct arrangements where person A gets can #1 and person B gets can #2, versus person A getting can #2 and Person B getting can #1? If the cans are identical, there is just one such possibility: persons A and B each get one can.

Yes the cans are identical.

My question is, say you have 3 identical cans, and 3 distinguishable people.

How many ways can I distribute these 3 cans to the 3 people. A person doesn't have to have any, or can have all of the cans.

So the possibilities are: Person A gets 0-3 cans, Person B get 0-3 cans, and Person C gets 0-3 cans.

I don't see how this would work with permutations/combinations because there will be a 0 in the product

ex: 3*2*1*0

...

edit:

it seems I have found a chapter in my book... "combinations with repetitions" that will possibly help me answer the question.

I am going to mark the thread as solved for now. If I don't figure it out do I just make it unsolved again and bump it? thanks.

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After you hand out the first can, there are no restrictions on handing out the second can. Three places to put the first can, and three places to put the second can. 3^2 possible choices. If the cans are distinguishable you are done. If the cans are indistinguishable you have to remove the double counting of the degenerate cases. There is only one way for anyone person to get both cans, so three cases are not double counted. The remaining 6 get reduced to three for a total of six.

To summarize, if there is no restriction on how many items anyone person can have, then there are M^N ways to distribute N distinguishable items to M people. If the items are indistinguishable, the accounting gets messy particularly for N > 2.

Yes the cans are identical.

My question is, say you have 3 identical cans, and 3 distinguishable people.

How many ways can I distribute these 3 cans to the 3 people. A person doesn't have to have any, or can have all of the cans.

So the possibilities are: Person A gets 0-3 cans, Person B get 0-3 cans, and Person C gets 0-3 cans.

I don't see how this would work with permutations/combinations because there will be a 0 in the product

ex: 3*2*1*0

...

edit:

it seems I have found a chapter in my book... "combinations with repetitions" that will possibly help me answer the question.

I am going to mark the thread as solved for now. If I don't figure it out do I just make it unsolved again and bump it? thanks.

The number you want is equal to the number of non-negative integer solutions of the equation ##x_A + x_B + x_C = 2.## By looking separately at the cases ##x_A = 0, x_A = 1## and ##x_A = 2## you can very rapidly solve the problem, to get 6.

If you have ##n## identical cans, the number of different way to distribute them to A, B and C is the number of non-negative integer solutions to the equation ##x_A + x_B + x_C = n.## Again, by looking at the separate cases of ##x_A = 0, 1, 2, \ldots, n## you reduce the problem to a series of simpler problems. If we fix ##x_A = k## we then need to find the number ##N_{BC}(n-k)## of ways of distribution ##n-k## cans to B and C. If you look at that problem graphically (in a two-dimensional plot in ##(x_B, x_C)## space) you will soon see the formula for ##N_{BC}(n-k).## Now just sum over ##k.##

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