9.18 Solve the system of equations

$$
\left\{\begin{array}{l}
A_{y}^{x}: P_{x-1}+C_{y}^{y-x}=126 \\
P_{x+1}=720
\end{array}\right.
$$
9.18 From the second equation, it follows that $(x+1)!=720$; since $720=6!$, then $x=5$. Considering that $C_{y}^{y-x}=C_{y}^{x}$ (according to formula (9.6)), we rewrite the first equation as: $A_{y}^{5}: P_{4}+C_{y}^{5}=126$. Further, using formula (9.4), we have $A_{y}^{5}: P_{4}=P_{5} C_{y}^{5}: P_{4}=5 C_{y}^{5}$, from which we obtain the equation $6 C_{y}^{5}=126$, or $y(y-1)(y-2)(y-3)(y-4)=21 \cdot 120$. But $21 \cdot 120=2^{3} \cdot 3^{2} \cdot 5 \cdot 7=7 \cdot 6 \cdot 5 \cdot 4 \cdot 3$, from which $y=7$.

Answer: $(5 ; 7)$.