help

Just count them:

1   = R 0
1   1   = R 1
1   2   1   = R 2
1   3   3   1   = R 3
1   4   6   4   1   = R 4
1   5   10   10   5   1   = R 5
1   6   15   20   15   6   1   = R 6
1   7   21   35   35   21   7   1   = R 7
1   8   28   56   70   56   28   8   1   = R 8
1   9   36   84   126   126   84   36   9   1   = R 9
1   10   45   120   210   252   210   120   45   10   1   = R 10

Thanks Guest that is a good answer.

I just counted them too but  I did not need to work out all the numbers.

I drew it in the shape of the normal triangle but insead of putting the actual number I just put  o  for odd and  e  for even.

If 2 odds are added the answer will be even

If 2 evens are added the answer will be even 

If one even and one odd are added the answer will be odd.

n choose k is NOT divisible by some prime number p if and only if:

\(\{\frac{k}{p^i}\} \leq \{\frac{n}{p^i}\} \) ({x} is the fractional part of x) for all \(i\in\mathbb{N}\) 

In other words, if the representation of n in base p is a_ma_m-1....a₀ and the representation of k in base p is b_mb_m-1....b₀ then n choose k is NOT divisible by p if and only if \(\forall 0\leq i \leq m\enspace b_i \leq a_i\)

conclusion: if the representation in base p of n is a_ma_m-1....a₀ then there are exactly \(\prod_{i=0}^{m} (a_i+1)\) values of k between 0 and n such that n choose k is NOT divisible by p

EDIT: here's a useful stack exchange thread