What is $1000 \choose 326$ $\mod{13}$? $1000=5(13^2)+11(13)+12$
$326=1(13^2)+12(13)+1$
Now, using Lucas theorem we have
$1000 \choose 326$$\equiv$$5 \choose 1$$11 \choose 12$$12 \choose 1$ $\pmod{13}$
Here, we got into the trouble since $11 \choose 12$ is not defined.
So, how do we compute this?

The exponent of 13 on the factorisation of $1000!$ is $\lfloor\frac{1000}{13}\rfloor+\lfloor\frac{1000}{13^2}\rfloor$ do the same for $326!$ and $674!$ and you'll find that after dividing the exponent on 13 will be greater than one, so the residue modulo 13 is 0.
