9.19 For what values of $x$ and $y$ is the equality possible

$$
C_{y}^{x}: C_{y+2}^{x}: A_{y}^{x}=1: 3: 24 ?
$$
9.19 Applying formula (9.4), we get

$$
\frac{y!}{x!(y-x)!}: \frac{(y+2)!}{x!(y-x+2)!}=\frac{1}{3} ; C_{y}^{x}:\left(x!C_{y}^{x}\right)=\frac{1}{24}
$$

From the second equation, we have $x!=24$, i.e., $x=4$ (since $24=1 \cdot 2 \cdot 3 \cdot 4$), and from the first equation, we find

$$
\frac{(y-x+1)(y-x+2)}{(y+1)(y+2)}=\frac{1}{3}
$$

Since $x=4$, we have $y^{2}-9 y+8=0$, from which $y=1$ and $y=8$; $y=1$ does not satisfy the condition (it should be $y>x=4$). Therefore, the answer is $(4 ; 8)$.