Let $p_k(n)$ denote the number of partitions of $n$ into exactly $k$ parts. It is known that $p_k(n)$ satisfies the recurrence
$p_k(n) = p_{k-1}(n-1) + p_k(n-k)$,
where $p_k(n)=0$ for $k>n$, $p_n(n)=1$, and $p_1(n)=1$. It is also known that there are $\binom{n-1}{k-1}$ compositions of $n$ into $k$ parts. Clearly, $p_k(n) \leq \binom{n-1}{k-1}$. Let $c(n,k)>0$ be a positive constant such that $c(n,k)\cdot p_k(n) \leq \binom{n-1}{k-1}$.
What are known upper bounds for the value of $c(n,k)$? I'm looking for a tight upper bound such that $c(n,k)\cdot p_k(n)$ does not exceed $\binom{n-1}{k-1}$.
I have read about upper bounds for $p_k(n)$. However, some of these are greater than $\binom{n-1}{k-1}$, which is not desirable.
EDIT: early replies of $k!$ led me to edit this question.
The upper bound for the ratio is $k!$, which is approached as $n$ increases without limit.
For example:
$k!$ is a an upper bound because you can order the $k$ parts of the partition in no more than $k!$ different ways to make different compositions; exactly $k!$ ways if all the parts are distinct but fewer if some are equal.
For large enough $n$ given $k$, the proportion of partitions with equal parts can be arbitrarily small. See this by considering $q_k(n)$ as the number of partitions of $n$ into $k$ distinct positive parts; then $q_k(n)= p_k\left(n-\frac{k(k-1)}{2}\right)$ so $\frac{q_k(n)}{p_k(n)} \to 1$ as $n$ increases