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.

#Combinatorics

Easy Math Editor

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Comments

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.

Brian Charlesworth - 5 years, 1 month ago

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 - 5 years, 1 month ago

It also depends if the boxes can be empty or not

Vignesh S - 5 years, 1 month ago

Is it $66$ ?

Vignesh S - 5 years, 1 month ago

They can be empty.

Adarsh Kumar - 5 years, 1 month ago

If they can be empty then it has to be 91

Vignesh S - 5 years, 1 month ago

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

Adarsh Kumar - 5 years, 1 month ago

@Adarsh Kumar – Sure. No problem

Vignesh S - 5 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$