Candy Bar Problem: distribute among three people

In how many ways can you distribute 9 identical candy bars to 3 people if each person does not have to recieve any candy bars?

There is a rather advanced concept of counting the number of ways to place N identical objects into K different cells: here the candies are the objects and the people are the cells.
\(\displaystyle \L \left( {\begin{array}{c}
{N + K - 1} \\
N \\
\end{array}} \right) = \frac{{\left( {N + K - 1} \right)!}}{{N!\left( {K - 1} \right)!}}\)

You posted in Beginning Algebra. This seems a bit to advanced for that level.

I am sorry. The problem sounded easy, but I see know that there is good reason why I was blank when looking at it

Could you explain that equation a little more. I was looking at something called permutations i think... the nCr and nPr. does that have something associated with this problem?

As I said before, this is advanced as you can now see for yourself.

Math wiz ya rite 09 said:

Could you explain that equation a little more. I was looking at something called permutations i think... the nCr and nPr. does that have something associated with this problem?

Click to expand...

I have never encountered a question exactly like this, however now I have an intuitive understanding from pka's example. Not to be redundant, but if you are still confused I will try and explain why it is true:

"High-School"-level (or beginning college) probability/statistics gives us the following result: If w is a word that has n letters consisting of two diferent characters a and b in which a appears j times and b appears k times then there are \(\displaystyle \frac{n!}{j!k!}\) ways to form a distinct "word" from the given word. Since \(\displaystyle n=j+k\) we have \(\displaystyle \frac{(j+k)!}{j!k!}\). This also generalizes to larger words with more distinct letters. For example, the number of ways to rearrange "Baboon" are \(\displaystyle \frac{6!}{2!2!1!1!} = 180\) where 2,2,1,1 represent the repetitions of b's, o's a's and n's respectfully.

Using the diagram/info pka gave,

If there are N objects and K holes then we have N+(K-1) "letters". Why? Because in the example pka posted with N=5,K=3, we need two separators to form the three boxes. (Analogously, we need to draw only two vertical lines to form the three columns of a tic-tac-toe board.) So (K-1) separators mimic the K holes, and along with the N identical objects we have N+(K-1) things to permute. But in permuting them, we have N identical objects (i.e. "letters"), and (K-1) identical separators (i.e. also "letters").

Using our formula we have \(\displaystyle \frac{\(N+(K-1)\)!}{N!(K-1)!}\).

Hope that made sense,
-daon