Probability doubt!

If one has to place 12 identical balls in 3 different boxes,is the number of ways 3123^{12} or the number of non-negative integral solutions to the equation a+b+c=12a+b+c=12?

I think it is the latter one,since if we are taking the product, we are assuming that the balls are distinct. Please provide insight.

Note by Adarsh Kumar
3 years, 5 months ago

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You are correct to think that it is the latter. Since the balls are identical, all that matters is the number of non-negative integer solutions (a,b,c)(a,b,c) to the equation you mention, the answer being (12+3131)=91\binom{12 + 3 - 1}{3 - 1} = 91.

If the balls are distinct and the order that they are placed in the boxes doesn't matter then the answer is 3123^{12}. If the order does matter, (e.g., if a solution where ball 1 is positioned under ball 2 is different than a solution with ball 2 positioned under ball 1), then things get more complicated.

Brian Charlesworth - 3 years, 5 months ago

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Thanks a lot sir!Actually this problem is from an exam(quite popular in India),and the options contained nothing like 91,so..

Adarsh Kumar - 3 years, 5 months ago

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It also depends if the boxes can be empty or not

Vignesh S - 3 years, 5 months ago

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Is it 6666?

Vignesh S - 3 years, 5 months ago

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They can be empty.

Adarsh Kumar - 3 years, 5 months ago

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If they can be empty then it has to be 91

Vignesh S - 3 years, 5 months ago

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@Vignesh S Yup!Actually in the book it was written as 3123^{12},that is why I posted this note.Thanx!

Adarsh Kumar - 3 years, 5 months ago

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@Adarsh Kumar Sure. No problem

Vignesh S - 3 years, 5 months ago

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