# In what circumstance do we have $k \geq \frac{{n \choose 6}}{{n \choose 3}}$ implies $k \geq n^3$?

In one of my lectures we derive the inequality $$k \geq \frac{{n \choose 6}}{{n \choose 3}}$$, where $$k$$ is a variable and $$n$$ the number of vertices in a graph. The lecturer then proceeds to state that "for $$n$$ large enough, we have $$k \geq n^3$$". I can see that $$\frac{{n \choose 6}}{{n \choose 3}} = \frac{(n-3)(n-4)(n-5)}{6\cdot 5\cdot 4}$$, and as $$n$$ grows large, $$n^3 \approx (n-3)(n-4)(n-5)$$. But the RHS never grows bigger than the LHS. So how can we make that claim about $$k$$?

You can't. What you can say, and what the lecturer probably meant, is that for $$n$$ large enough we have $$k\geq cn^3$$ where $$c>0$$ is some absolute constant. Presumably this is good enough for whatever you actually use the bound to prove.
• Yes, the question is actually about the existence of the constant $c$ and the integer $n^*$ such that for $n \geq n^*$, certain condition involved $k$ is satisfied. I was thinking along the line of "finding $c$", but it seems more like $c$ is something we would specify, and then we find the appropriate $n^*$ to satisfy the condition on $k$? – ensbana Dec 24 '19 at 10:59
• @ensbana yes, you could choose any value of $c$ which is strictly less than $1/120$ and then calculate an explicit $n^*$ if you wanted to. (For example, if you pick $c=1/200$ then $n^*$ works out to be $26$.) – Especially Lime Dec 24 '19 at 11:11