×

# 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.

7 months, 2 weeks ago

Sort by:

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. · 7 months, 2 weeks ago

Thanks a lot sir!Actually this problem is from an exam(quite popular in India),and the options contained nothing like 91,so.. · 7 months, 2 weeks ago

It also depends if the boxes can be empty or not · 7 months, 2 weeks ago

They can be empty. · 7 months, 2 weeks ago

If they can be empty then it has to be 91 · 7 months, 2 weeks ago

Yup!Actually in the book it was written as $$3^{12}$$,that is why I posted this note.Thanx! · 7 months, 2 weeks ago

Sure. No problem · 7 months, 2 weeks ago

Is it $$66$$? · 7 months, 2 weeks ago