(1) A recursion $p_k(n)=\sum_{m=1}^n p_{k-1}^{s m}(n-m)$, where we have sorted partitions of $n$ with $k$ parts by their maximum part, and then deleted that part. Here $p_{k-1}^{\leq m}(n-m)$ is the number of partitions of $n-m$ into exactly $k-1$ parts, all of which are at most $m$. Explicitly, there is a bijection $ \bigcup_{m=1}^n\{\text { partitions of } n-m \text { with } k-1 \text { parts all } \leq m\} \rightarrow\{\text { partitions of } n \text { with } k \text { parts }\} $ which, given a choice of $m$ and a partition $Q$ of $n-m$ with $k-1$ parts all $\leq m$, outputs the partition $Q+m$ of $n$. (Here $Q+m$ means "append $+m$ to Q.) The inverse bijection, applied to a partition $Q$ of $n$ with $k$ parts, deletes the largest part of $Q$. (2) A recursion $p_k(n)=p_{k-1}(n-1)+p_k(n-k)$, where we have sorted partitions of $n$ with $k$ parts into two types: those that contain at least one ' 1 ', and those that do not. For the former, we delete the ' 1 ' to get a partition of $n-1$ with $k-1$ parts, and for the latter, we subtract 1 from every part to get a partition of $n-k$ with $k$ parts. Explicitly, there is a bijection \{partitions of $n-1$ with $k-1$ parts\} $\cup\{$ partitions of $n-k$ with $k$ parts $\} \rightarrow\{$ partitions of $n$ with $k$ parts \} that sends a partition $Q$ of $n-1$ with $k-1$ parts to $Q+1$, and sends a partition $Q$ of $n-k$ with $k$ parts to the partition of $n$ obtained by increasing each part by 1 . The inverse bijection takes a partition $Q$ of $n$ into $k$ parts, and deletes a ' 1 ' from $Q$ if $Q$ contains any ' 1 's, and otherwise decreases each part of $Q$ by 1 . (3) We also used a stars/bars argument to count "ordered partitions" of $n$ into $k$ parts, where e.g $1+4$ and $4+1$ are counted as different partitions of 5 . We argued that the number of these is the same as the number of ways to divide $k$ cookies among $n$ people, where everyone gets at least 1 cookie. We then argued that this number is $\binom{k-1}{n-1}$.