The fastest way is probably to use the formula, and not pascals triangle. Let's start not to do multiplications when we know that we're going to divide by the same number later.
If k < n/2, let's have k = n - k. We know that C(n,k) = C(n,n-k)
Now :
n! / (k! x (n-k)!) = (product of numbers between (k+1) and n) / (n-k)!
At least with this technique, you're never dividing by a number that you used to multiply before. You have (n-k) multiplications, and (n-k) divisions.
I'm thinking about a way to avoid all divisions, by finding GCDs between the numbers that we have to multiply, and those we have to divide. I'll try to edit later.