$4 \cdot 30 \quad a>b>c, \quad x>y>z, \quad M=a x+b y+c z, \quad N=a z$ $+b y+c x, \quad P=a y+b z+c x, \quad Q=a z+b x+c y$. Then we have
(A) $M>P>N$ and $M>Q>N$.
(B) $N>P>M$ and $N>Q>M$.
(C) $P>M>Q$ and $P>N>Q$.
(D) $Q>M>P$ and $Q>N>P$.
(1st "Hope Cup" National Mathematics Invitational Competition, 1990)
[Solution] $M-N=a(x-z)+c(z-x)=(a-c)(x-z)>0$,
then $M>N$;
$$
N-P=a(z-y)+b(y-z)=(a-b)(z-y)>0 \text {, }
$$
(notice $a>b>c, \quad x>y>z$ )
$$
\therefore \quad M>P \text {. }
$$

Similarly, we can prove $M>Q$.
In summary, $M>P>N$, and $M>Q>N$.
Therefore, the answer is $(A)$.