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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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. – Brian Charlesworth · 9 months ago

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– Adarsh Kumar · 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..Log in to reply

It also depends if the boxes can be empty or not – Vignesh S · 9 months ago

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– Adarsh Kumar · 9 months ago

They can be empty.Log in to reply

– Vignesh S · 9 months ago

If they can be empty then it has to be 91Log in to reply

– Adarsh Kumar · 9 months ago

Yup!Actually in the book it was written as \(3^{12}\),that is why I posted this note.Thanx!Log in to reply

– Vignesh S · 9 months ago

Sure. No problemLog in to reply

– Vignesh S · 9 months ago

Is it \(66\)?Log in to reply