Note 2188=2^2/547 so by legendre's formula, we can compute v2(2188!) ,and v547(2188!) is clearly 4. We need the number of k's such that v2(k!(2188-k)!)<v2(2188!) and v547(k!(2188-k)!)<=3

We notice that $547$ divides $\binom{2188}{k}$ if and only if $k$ is not a multiple of $547$ so $k$ cant be $547$, $547\times2$, $547\times3$, $2188$. Now we consider $V_2(\binom{2188}{k})$. $V_2(\binom{2188}{k})>2$ as $V_2(2188)=2$ The binary representation of $2188$ is $100010001100$ Using Kummer's Theorem, we know that if $V_2(\binom{2188}{k})=0$, there are $2^4$ ks If $V_2(\binom{2188}{k})=1$, $2^3\times2=16$ ks 2189-16-16-4+1=2154 numbers

Essentially equivalent to CGMO 2012/8, just $2188$ in place of $2012$.