# Probability doubt!

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. 3 years, 9 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 $(a,b,c)$ 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.

- 3 years, 9 months ago

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

- 3 years, 9 months ago

It also depends if the boxes can be empty or not

- 3 years, 9 months ago

Is it $66$?

- 3 years, 9 months ago

They can be empty.

- 3 years, 9 months ago

If they can be empty then it has to be 91

- 3 years, 9 months ago

Yup!Actually in the book it was written as $3^{12}$,that is why I posted this note.Thanx!

- 3 years, 9 months ago

Sure. No problem

- 3 years, 9 months ago