jmc

If the row contains a $1$, then a $6$, then the binomial coefficients must be $\left(\genfrac{}{}{0ex}{}{6}{0}\right)$ and $\left(\genfrac{}{}{0ex}{}{6}{1}\right)$. All we need to check now are $\left(\genfrac{}{}{0ex}{}{6}{2}\right)$ and $\left(\genfrac{}{}{0ex}{}{6}{3}\right)$, since $\left(\genfrac{}{}{0ex}{}{6}{0}\right)=\left(\genfrac{}{}{0ex}{}{6}{6}\right)$, $\left(\genfrac{}{}{0ex}{}{6}{1}\right)=\left(\genfrac{}{}{0ex}{}{6}{5}\right)$ , and $\left(\genfrac{}{}{0ex}{}{6}{2}\right)=\left(\genfrac{}{}{0ex}{}{6}{4}\right)$. $\left(\genfrac{}{}{0ex}{}{6}{2}\right)=\frac{6!}{4!\times 2!}=15$, and $\left(\genfrac{}{}{0ex}{}{6}{3}\right)=\frac{6!}{3!\times 3!}=20$. None of those is prime, so there are $\boxed{0}$ prime numbers in the given row.