Number of unordered 4 partitions of 12, unordered $k$ partitions of natural number $n$:

Question

I just ended up enumerating which was time consuming and I'm sure it's not what I was supposed to do $$1119\quad 2217\quad 3333\quad 1425\\ 1128\quad 2226\quad 3342\quad 1236\\ 1137\quad 2235\quad 3351\quad 1434\\ 1146\quad 2244\quad 1155\quad$$ So I guess there are 15. Is there an easier combinatorial way to do this?

Answer 1

You can use the following recursive formula. Let $P_k(n)$ be the set of partitions of $n$ with exactly $k$ parts. Let $p_{k}(n)$ be the number of partitions of $n$ into $k$ parts i.e. $p_{k}(n)=|P_k(n)|$. Then $$ p_{k}(n)=p_{k-1}(n-1)+p_{k}(n-k) $$ by classifying $\lambda=(\lambda_1,\dots,\lambda_k)\in P_k(n)$ (where the $\lambda_i$ are weakly decreasing and positive) based on whether $\lambda_k=1$ or $\lambda_k>1$.