Answer: B

Solution:

A quadratic fails to have two distinct real solutions exactly when its discriminant is nonpositive. Thus we need $$b^2-4c\le0\qquad c^2-4b\le0,$$ 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 $$\begin{gathered} (c=1,2),\quad (c=1,2), \\ \quad (c=3),\quad (c=4). \end{gathered}$$

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

Thus, B is the correct answer.