# Calculating the number of possible outcomes

• October 12th 2010, 02:25 AM
ah81
Calculating the number of possible outcomes
Hi everyone,

Now, when it comes to all things maths, I am totally hopeless. I have lots of trouble solving basic problems, so I knew I'd have to turn to the web for help with my current problem. Forgive the lay terms I use, it's the only way I can express things!

OK, imagine you have a small box with 9 smaller compartments (a 3x3 grid) and 9 pebbles that can be placed inside each of the compartments. I'm trying to figure out how many different combinations of pebbles are possible.

Attachment 19286

In the above image, you can see that the first row of the grid is filled in the first, second and third columns. Then, a few different combinations of rows and columns.

So, I'm trying to figure out how many possible combinations there are available (from 0 - 9 pebbles and everything in between), and what the best way to figure out those combinations is.

As I said, I'm not good with this stuff, but any tips or formulas that anyone can offer me would be greatly appreciated!
• October 12th 2010, 07:13 AM
emakarov
If I understand right, you need the number of ways to place every number of pebbles between 0 and 9 into every combination of compartments. Then you can reason as follows. The first compartment can be filled or not -- 2 possibilities. For each of those two, the second compartment cab be filled or not -- total four possibilities. For each of those four, the third can be filled or not -- this gives eight possibilities, etc. The total number is $2^9$ possibilities.
• October 12th 2010, 07:39 AM
Plato
I did not answer the OP earlier because I did not think the question was clear. $2^9$ is the number of subsets of a nine element set. Is that what is being asked? I don’t think so. I think the object is to actually put all nine pebbles into the box. If we put $\boxed{2}\boxed{4} \boxed{3}$ across the top, that is different from $\boxed{3}\boxed{2} \boxed{4}$.
Moreover $2^9$ counts an empty box.

If we take it to mean that the nine cells are labeled and the pebbles are identical then the answer is $\dbinom{9+9-1}{9}=\dfrac{17!}{9!\cdot 8!}$.

If we take it to mean that the nine cells are labeled and the pebbles are also labeled then the answer is $9^9$.

I wonder if rotations of the grid should be considered, particularly if neither is labeled.
• October 12th 2010, 08:33 AM
emakarov
So, the OP should clarify (1) whether pebbles are labeled or identical, (2) whether more than one pebble can be placed in a cell, and (3) whether leaving all cells empty is a legitimate choice.
• October 12th 2010, 03:44 PM
ah81
Sorry for the vague nature of my question guys. I'll see if I can make it a little clearer.

To answer your questions emakarov, the pebbles are all identical, only one can be placed in a cell and leaving all cells empty is a legitimate choice. As you stated in your first post, it may help to think of each cell as filled or not filled.

The image below shows the path I'm on a little clearer...

http://i160.photobucket.com/albums/t...hill/grid2.jpg

I'm sure you can see what I'm getting at. 0 & 1 pebbles were easy, as was 8 & 9. It's 2 - 7 that obviously yield much more permutations, and I'm sure I would go loopy trying to figure it out myself!

Hope this makes my question a little more easy to understand.
• October 12th 2010, 04:00 PM
Plato
Quote:

Originally Posted by ah81
0 & 1 pebbles were easy, as was 8 & 9. It's 2 - 7 that obviously yield much more permutations, and I'm sure I would go loopy trying to figure it out myself!

Thank you for that replay. While it clarifies what you want, it shows some confusion.
There is only one way to put no pebbles into the box: not 8.
There are nine ways to put one pebble into the box.
Select two of the cells, $\binom{9}{2}=\frac{9\cdot 8}{2}=36$ ways to put two pebbles into the box.
So the total number of ways is indeed $2^9=512$.
• October 12th 2010, 11:15 PM
ah81
Quote:

Originally Posted by Plato
There is only one way to put no pebbles into the box: not 8.

Sorry Plato, what I meant that it was easy to calculate the combinations for 0, 1, 8 & 9 pebbles.

Thanks for the formulas, it looks to me though that there are going to be quite a lot of possible combinations that can be created, more than I can be bothered working out!