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

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.

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## Comments

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TopNewestYou 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 \((a,b,c)\) to the equation you mention, the answer being \(\binom{12 + 3 - 1}{3 - 1} = 91\).

If the balls are distinct and the order that they are placed in the boxes doesn't matter then the answer is \(3^{12}\). 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.

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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..

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It also depends if the boxes can be empty or not

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Is it \(66\)?

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They can be empty.

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If they can be empty then it has to be 91

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