How many ways are there to distribute 12 unlabeled b***s into 9 labeled boxes

Here's another way of looking at this:

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What is it with all the box and ball questions? Have the new text books arrived or are we are all in a billiards tournament getting our b***s racked?

We’ll find out after the break . . .

$$\displaystyle \left( {\begin{array}{*{20}c} 11 \\ 8 \\ \end{array}} \right)\; = \; \dfrac{11!}{8!(11-8)!} \; = \;\hspace{5pt} \Text {165 \ ways} \\\$$

WHY is this the answer Nauseated?  Can you please explain?

Thanks, Alan....that explanation is very clear......

Nauseated......could you explain your answer???.....like Melody, I'm curious to see how you arrived at that......(even though it's correct)

I am playing with this without really looking at the other answers - or if I did I have forgotten.

I just want to see if i get the same answer.  :/

12 identical b***s

9 boxes

at least one ball in each box

so really the problem is just the same as 3 identical b***s in 9 boxes

All in the same box that is 9 ways

2 in one box and 1 in the other  9*8 = 72                or      2*9C2=72

all in different boxes maybe that is  9C3 = 84 ways

Total number of ways is    9+72+84 = 165

WOW I got the same answer as Alan and Nauseated.      ヽ(•‿•)ノ

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Now Alan I do not understand your sigma notation.  Could you talk about that please  :/

And Nauseated.  I still don't get where your formula came from either, did you just pluck it off the magic tree, like a golden apple?   :/

Sigma notation explanation:

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Thank you for explaining the double sigma notations Alan      (I didn't understand that before)

BUT

I did this with 9C3   Choose three boxes our of 9 and put an identical ball in each one.  9C3 = 84

Now,  your summation gives the same result but I don't understand the logic that you used to derive the sum in the first place.  (where on Earth did you pluck 7 from ??? )

Can you try and explain the logic that you used please?

Think of the first summation as the position of the first ball.  It can only go into boxes 1 to 7, because the 2nd ball will be in another box to the "right" of it, and the third ball will be to the "right " of that.

It might have been better to write the summations as follows, where each summation refers to each ball:

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Thanks Alan,

I will chew on this when I am less tired :)

I am sorry Alan, I just don't get it.

If I want to know how many ways I can put 3 identical b***s 9 different boxes such that there was a maximum of 1 ball in each box then I would say

1.       9C3=84        or

2.       $$\frac{9*8*7}{3!}=84$$

Now , you have said

So               $$1+3+6+10+15+21+28 =$$   $$\frac{9*8*7}{3!}=84$$

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OR

I am really struggling with this Alan.   

I totally understand that the answer to this is 84 (doing it my way) but I just can't understand the logic that you use to do it your way.

ALSO  can you show me how to expand this triple summation please.  I have no idea :(
I have always had trouble with sigma notation.  As soon as it is above the most basic level I get lost :/

Maybe you could throw me a couple of double summations that I could try on my own. :)

Here is an explanation of the double summation and a couple of practice double summations Melody:

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Confusion with formulas

How many ways are there to distribute 12 unlabeled b***s into 9 labeled boxes in such a way that each box receives at least one ball? 

(12+9-1)C12

$${\left({\frac{{\mathtt{20}}{!}}{{\mathtt{12}}{!}{\mathtt{\,\times\,}}({\mathtt{20}}{\mathtt{\,-\,}}{\mathtt{12}}){!}}}\right)} = {\mathtt{125\,970}}$$

Mmm that is not right      

ok

(12-1)C(9-1)      this was the formula Nauseated used and I have seen it before but I am confused

When do you use this formula     (n-1)C(k-1)

and when do you use    (n+k-1)C(n)