$4 . a 、 b 、 c 、 d$ are pairwise distinct positive integers, and $a+b=c d, a b=c+d$. Then the number of quadruples $a 、 b 、 c 、 d$ that satisfy the above requirements is $(\quad)$.
(A) 4
(B) 6
(C) 8
(D) 10
4. (C).
(1) If $a, b, c, d$ are all not equal to 1, and $a \neq b, c \neq d$, then $(a-1)(b-1) \geqslant 2$, i.e., $a b > a + b$, $(c-1)(d-1) \geqslant 2$, i.e., $c d > c + d$.
But $a + b = c d > c + d, c + d = a b > a + b$, which is a contradiction. Therefore, there is no solution in this case.
(2) If one of $a, b, c, d$ is 1, without loss of generality, let $a=1$, then
$$
\begin{array}{l}
\left\{\begin{array}{l}
1 + b = c d, \\
b = c + d
\end{array} \Rightarrow c d = c + d + 1\right. \\
\Rightarrow (c-1)(d-1) = 2 \\
\Rightarrow\left\{\begin{array}{l}
c - 1 = 1, \\
d - 1 = 2
\end{array} \text { or } \left\{\begin{array}{l}
c - 1 = 2, \\
d - 1 = 1,
\end{array}\right.\right.
\end{array}
$$

i.e., $\left\{\begin{array}{l}c=2, \\ d=3\end{array}\right.$ or $\left\{\begin{array}{l}c=3, \\ d=2 \text {. }\end{array}\right.$ In this case, $b=5$.
Similarly, when $b=1, c=1, d=1$, we can get 6 more solutions.
Therefore, there are a total of 8 solutions:
$$
\begin{array}{l}
(1,5,2,3),(1,5,3,2),(5,1,2,3),(5,1,3,2), \\
(2,3,1,5),(2,3,5,1),(3,2,1,5),(3,2,5,1) .
\end{array}
$$