I am interested in counting the number of ways of partitioning N integers (N is a multiple of 3) to N/3 triples. I came up with this formula for which I am looking for a closed form or tight asymptotic upper-bound:
$C = \prod_{j=0}^{j=(N-3)/3} \binom{N-3j}{3} $
Trivial upper-bound is $N^N$. I am looking for tight upper-bound.
On
Denoting the number of triples as $n$ one obtains: $$ C=\frac{(3n)!}{n!(3!)^n}. $$ Then apply Stirling's formula to obtain $C\sim\sqrt{3}\left(\frac{9n^2}{2e^2}\right)^n$.
Define $n=\frac N3$ as the number of triplets. If you don't care about the order the triplets are selected, the number of ways is $$\frac {(3n)!}{(3!)^nn!}$$ You can get this by permuting the numbers in one of $(3n)!$ ways and taking the first three as the first triplet, the next three as the next triplet, and so on. We don't care about the order of the three in each triplet, which gives us $n$ divisions by $3!$. We also don't care about the order of the triplets, so we divide by $n!$ Applying Stirling's approximation to the large factorials we get $$\frac {(3n)!}{(3!)^nn!}\approx \frac {(3n)^{3n}e^n\sqrt{2\pi3n}}{e^{3n}6^nn^n\sqrt{2\pi n}}=\sqrt 3\frac{n^{2n}27^n}{e^{2n}6^n}\approx \sqrt 3\ \cdot0.609^nn^{2n}$$