[m]С^{y+1}_{x+1}=\frac{(x+1)!}{(x+1–(y+1))!\cdot (y+1)!}=\frac{(x+1)!}{(x–y)!\cdot (y+1)!}[/m]
[m]С^{y}_{x+1}=\frac{(x+1)!}{(x+1-y)!\cdot y!}[/m]
[m]С^{y+2}_{x+1}:С^{y+1}_{x+1}=\frac{(x+1)!}{(x-y-1)!\cdot (y+2)!}:\frac{(x+1)!}{(x–y))!\cdot (y+1)!}=\frac{(x+1)!}{(x-y-1)!\cdot (y+2)!}\cdot\frac{(x–y)!\cdot (y+1)!}{(x+1)!}=\frac{(x–y-1)!\cdot (x-y)\cdot (y+1)!}{(x-y-1)!\cdot (y+1)!\cdot (y+2)}=\frac{ x-y}{y+2}[/m]
[m]С^{y+1}_{x+1}:С^{y}_{x+1}=\frac{(x+1)!}{(x+1–y-1)!\cdot (y+1)!}:\frac{(x+1)!}{(x+1-y)!\cdot y!}=\frac{(x+1)!}{(x–y)!\cdot (y+1)!}\cdot \frac{(x-y+1)!\cdot y!}{(x+1)!}=\frac{x-y+1}{y+1}[/m]
По условию:
[m]С^{y+2}_{x+1}:С^{y+1}_{x+1}:С^{y}_{x+1}==28:12:3[/m]
Поэтому
[m]С^{y+2}_{x+1}:С^{y+1}_{x+1}=28:12[/m] ⇒ [m]\frac{ x-y}{y+2}=\frac{28}{12}[/m] ⇒ [m]12(x-y)=28(y+2)[/m] ⇒ [m] 3x=10y+14[/m]
и
[m]С^{y+1}_{x+1}:С^{y}_{x+1}=12:3[/m] ⇒ [m]\frac{ x-y+1}{y+1}=\frac{12}{3}[/m] ⇒ [m]3(x-y+1)=12(y+1)[/m] ⇒ [m] 3x=15y+9[/m]
Решаем систему двух уравнений:
[m]\left\{\begin {matrix}3x=10y+14\\ 3x=15y+9\end {matrix}\right.[/m] ;
Приравниваем правые части
[m]\left\{\begin {matrix}3x=10y+14\\10y+14=15y+9\end {matrix}\right.[/m] ; [m]\left\{\begin {matrix}3x=10\cdot 1+14\\y=1\end {matrix}\right.[/m]
[m]\left\{\begin {matrix}x=8\\y=1\end {matrix}\right.[/m]