# Thread: Partition Theory

1. ## Partition Theory

In how many ways can a positive integer n be broken into a sum of k positive integers ? ( While representing the number as a sum the order in which the addends are arranged for each of the ways is not taken into consideration)

I know that we denote the number of ways to do this as $p(n,k)$ and I tried an inductive approach to the problem , i.e, I was able to figure out that ,

$p(n,1) = 1$

$p(n,2) = \left\lceil\frac{n-1}{2}\right\rceil$

$p(n,3) = \sum_{i=1}^{\left\lfloor\frac{n}{2}\right\rfloor} {\left\lceil\frac{i-1}{2}\right\rceil + \left\lceil\frac{n-i-1}{2}\right\rceil }$

But I cant seem to figure out $p(n,4)$ and so on....And I need a closed-form solution, not recurrence relations..So, please help . Thanks in advance .

2. ## Re: Partition Theory

Hello, mathlover14!

$\text{In how many ways can a positive integer }n$
$\text{be broken into a sum of }k\text{ positive integers?}$
$\text{(The order in which the addends are arranged is not considered.)}$

Note: The following does not answer the question.

If the order of the addends is considered, there is a simple approach.

Consider: $n = 8,\:k = 3$

Place the 8 objects in a row, leaving a space between them.
. . There are: $n-1\,=\,7$ spaces.

. . $\circ\;\_\;\circ\;\_\;\circ\;\_\;\circ\;\_\;\circ \;\_\; \circ\;\_ \;\circ\;\_\;\circ$

Select 2 of the 7 spaces and insert "dividers".
. . There are: $k-1\,=\,2$ dividers.

. . $\circ\;\_\;\circ\;\_\;\circ\;\_\;\circ\;|\;\circ \;|\; \circ\;\_ \;\circ\;\_\;\circ$

This diagram represents: $4 + 1 + 3 \:=\:8$

If the order is considered, the number of ways is: ${n-1\choose k-1}$

Can we eliminate the duplication included in this method?
I don't know . . . I've tried and failed.