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Let $n,p \in \mathbb N$ and consider the integer partition of $$\left\lfloor \frac{n(p-1)}{2} \right\rfloor$$ into $p$ or less parts, each of which is less or equal to $n-1$.

Can the number of partitions be bounded by $$ {m+p-1 \choose p}$$ with a suitable $m < n-1$?

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  • $\begingroup$ Intuitively, it feels like something like this should hold: the first expression is a set of Young diagrams, but a Young diagram can also be seen as a path, where one chooses up-steps or right-steps. The latter is easily counted by a binomial. $\endgroup$ – Per Alexandersson Oct 3 '14 at 12:15
  • $\begingroup$ I'd suggest to get rid of the "-1". In fact, in the answer of the other thread mathoverflow.net/questions/182401/…, it seems like n and p have been swapped by error. So you should just ask for the number of partitions of {np/2] with $\le p$ parts each $\le n$, to be bounded by $\binom {m+p}p$ for suitable m<n. $\endgroup$ – Wolfgang Oct 3 '14 at 15:38
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The number of partitions of $\lfloor np/2\rfloor$ with $\leq p$ parts each $\leq n$ is the middle coefficient of the $q$-binomial coefficient $\left[ n+p\atop p\right]$. For fixed $p$ it follows from http://math.mit.edu/~rstan/papers/qbc.pdf that this middle coefficient is asymptotic to $$ \frac{A(p-1,\lceil p/2\rceil)n^{p-1}}{(p-1)!p!} $$ as $n\to \infty$, where $A(p-1,\lceil p/2\rceil)$ denotes an Eulerian number. This seems to me to be a better result than bounding by ${m+p\choose p}$ for some $m$.

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Perhaps the closest you can get to the type of result you seek may be achieved with the following Erdős-Lehner result: As $N\rightarrow \infty$ the number of partitions of $N$ into exactly $M$ parts is asymptotic to $\frac{1}{M!}{N-1\choose M-1}$, where $M=o(n^{1/3})$.

See page 56 of GE Andrews, "The Theory of Partitions", Cambridge 1984.

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