C(n,r)

This is similar to an olympiad problem whose solution I had read before.

2010 = 2 \ \times \ 3 \times \ 5 \ \times 67

It is obvious that we must have n \ge 67

We need to consider only r≤n/2 (Since ⁿC_r=ⁿC_n-r)

Now if r≥2, then \binom {n}{r} \ge \binom {n}{2} \ge \binom {67}{2} = 2211>2010

Hence r =1 giving n=2010.

So the only solutions are (2010,1) and (2010,2009)