7.2. (Belgium, 79). Arrange the numbers

$$
\begin{gathered}
x=(a+b)(c+d) \\
y=(a+c)(b+d) \\
z=(a+d)(b+c)
\end{gathered}
$$

in ascending order, given that $a<b<c<d$.
7.2. Since $a0
$$

and

$$
z-y=a c+b d-a d-b c=(a-b)(c-d)>0
$$

Thus, $x<y<z$.