Gold hoarding groups

It's not really a generalization, but there's a way to use identities to solve it more efficiently. This kind of approach is helpful for creating a computer program to solve for partitions.

If $p(n)$ is the partition function of $n$ , and $p(n,r)$ is the partition function of $n$ into exactly $r$ parts, then:

$p(n)=\sum\limits_{k=1}^{n}{p(n,k)}$

Then, there's also the identities:

$p(n,1)=1$

p(n,2)=\left\{\begin{array}{111} \dfrac{n-1}{2} & , & \text{n is odd} \\ \dfrac{n}{2} & , & \text{n is even} \\ \end{array}\right.

$p(n,n)=1$

p(n,r)=\left\{\begin{array}{111} p(n-r) & , & n\le 2r \\ \sum\limits_{k=1}^{r}{p(n-r,k)} & , & n>2r \end{array}\right.

Now, using those identities:

$\begin{aligned} p(8)&=p(8,1)+p(8,2)+p(8,3)+p(8,4)+p(8,5)+p(8,6)+p(8,7)+p(8,8) \\ p(8)&=1+4+p(8,3)+p(4)+p(3)+p(2)+p(1)+1 \end{aligned}$

You can probably see how you can break it down even further in the next steps. There are more examples on the identical objects into identical bins page.

I think what you calculated was the number of ways to separate 8 distinct coins into only 2 groups. In this problem, note that the coins are identical . This means that it only matters how many coins are in each group; it doesn't matter which coins go in which group.

Here are a couple of ways to divide the coins into groups:

$4+4$

$4+2+2$

$3+2+2+1$

Also note that there can be any number of non-empty groups.

Another thing to keep in mind: There's an identity for what you are trying to do with the binomial coefficient:

$\sum\limits_{k=0}^{n}{\binom{n}{k}}=2^n$

Now, you left out $\binom{8}{0}=1$ in that calculation, so:

$\sum\limits_{k=1}^{n}{\binom{n}{k}}=2^n-1$

$\begin{aligned} \sum\limits_{k=1}^{8}{\binom{8}{k}}&=2^8-1 \\ &=255 \end{aligned}$