Solution:

A quadratic fails to have two distinct real solutions exactly when its discriminant is nonpositive. Thus we need $$b^2-4c\le0\quad\text{and}\quad c^2-4b\le0,$$ or $b^2\le4c$ and $c^2\le4b$.

From $b^2\le4c$, we get $b^4\le16c^2$. Combining this with $c^2\le4b$ gives $b^4\le64b$, so $b\le4$.

Now check $b=1,2,3,4$. The inequalities give respectively $$(c=1,2),\quad (c=1,2),\quad (c=3),\quad (c=4).$$

There are $2+2+1+1=6$ ordered pairs.

Thus, B is the correct answer.