how many combinations on a variable number of items?

I havent done combinations in forever, so I have no idea howto do this...

I have an unknown quantity of items in a set, and I need to figure out how many combinations there are of 35 unique items from that set. I know that the formula for combinations is $\frac{n!}{35!(n-35)!}$... but I need to express this as a polynomial for time complexity of an algorithm. is this doable?

To elaborate a bit on this:

I have to design an algorithm to show that a 35-Clique can be solved in polynomial time (k-clique is an exponential time complexity)... but knowing that there is a constant 35 nodes I need to search for supposedly puts this into polynomial territory, even though it'll still be a ridiculously high polynomial.

I found an example online using a 3-clique as a question, and it said that enumerating all triples $(u,v,w)$ of nodes in the graph would be a $O(|V|^3)$ complexity... But I don't see how $\displaystyle\frac{|V|!}{3!(|V|-3)!}$ became $|V|^3$...

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Exactly as in Gerry Myerson's answer $\displaystyle\frac{|V|!}{3!(|V|-3)!}=\frac{|V|(|V|-1)(|V|-2)}6$ then you ignore the denominator as you are interested in the power of the leading term. – Ross Millikan Feb 15 '13 at 5:12

The formula you have written reduces to the polynomial $${n(n-1)(n-2)\times\cdots\times(n-34)\over35!}$$

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You'll have to forgive my lack of understanding here.. An example I read online using an arbitrary k value converted the function above to $O(n^k)$, but I have no idea how this happened... – agent154 Feb 15 '13 at 5:00
If you imagine multiplying out the numerator (don't actually do it, just imagine doing it), you will get a polynomial with leading term $n^{35}$ and then terms of smaller degree. Do you understand why such a polynomial is $O(n^{35})$? – Gerry Myerson Feb 15 '13 at 5:19
OK... I see it now. Thanks – agent154 Feb 15 '13 at 5:21