Neither quadratic has two distinct real roots exactly when both discriminants are nonpositive: $b^2 \le 4c$ and $c^2 \le 4b.$

Combining $c\ge b^2/4$ with $c\le2\sqrt b$ gives $b^{3/2}\le8,$ so $b\le4.$ Checking: $b = 1$ gives $c \in \{1,2\};$ $b = 2$ gives $c \in \{1,2\};$ $b = 3$ gives $c = 3;$ and $b = 4$ gives $c = 4.$

That is $(1,1),$ $(1,2),$ $(2,1),$ $(2,2),$ $(3,3),$ $(4,4)$ — $6$ ordered pairs.

Thus, the correct answer is B.