# Partitions tending to a constant

$P_{k}(n)$ = the number of partitions of n into k parts. Now, if we fix some $t\ge 0$ , then $\lim_{n\to\infty}P_{n-t}(n)\to$ c, c being some constant.

Please help me with this! I believe $P_{k}(n)=\dbinom{n+k-1}{n}$. If that be so then, $P_{n-t}(n)$ comes out to some horrible combinatorial and I can't proceed any further to show that it'll be bounded by a constant. Also I'll be extremely glad if you could provide me more insight into the result.

• Hint: Expand and use Stirling's approximation of factorial en.wikipedia.org/wiki/Stirling's_approximation Commented Sep 13, 2014 at 20:45
First let me observe that the OP has the formula for compositions as opposed to partitions. Assuming the OP refers to partitions there is no need for calculus or Stirling's approximation. As we are considering the limit as $n$ goes to infinity we may assume that $n\ge 2t.$ Imagine the partition as a sequence of adjacent columns of stacked boxes, with the height of every stack being the value of the part. As we have $n-t$ parts $\ge 1$ we get a minimal contribution of $n-t$ from those parts and this is the bottom row. That leaves $t$ boxes to be distributed into the $n-t$ columns and since $n-t\ge t$ there are $P(t)$ configurations (themselves partitions) we can do so taking symmetry into account. That is our constant, it has the value $P(t).$ QED.
• Could you specify what OP is? ....I did come up with this argument but then I began to doubt it. How do I cover the cases there is only a stack filled and the rest remain empty! Then I need to consider the fact that these may permute among each other, but as n tend to infinity , then (seemingly) my values for $P_{n-t}(n)$ does too! Perhaps Stirling's form would now cover this case. Commented Sep 14, 2014 at 3:54
Hint: Expand $P_{n-t}(n)$ and use Stirling's approximation for factorial:
$$n! \sim \sqrt{2\pi n} \left( \frac{n}{e} \right)^n$$