Задача 54771 ...
Условие
Решение
[m]С^{y}_{x}С^{y}_{x}=\frac{x!}{(y)!\cdot (x-y)!};[/m]
[m]С^{y-1}_{x}=\frac{x!}{(y-1)!\cdot (x-y+1)!};[/m]
[m]С^{y+1}_{x}:С^{y}_{x}:С^{y-1}_{x}=6:14:21[/m] ⇒ запишем две пропорции:
[m]С^{y+1}_{x}:С^{y}_{x}=6:14[/m] ⇒[m] \frac{x!}{(y+1)!\cdot (x-y-1)!}: \frac{x!}{(y)!\cdot (x-y)!} =\frac{3}{7}[/m]
[m] \frac{x!}{y!\cdot (y+1)\cdot (x-y-1)!}\cdot \frac{y!\cdot (x-y-1)!\cdot (x-y)}{x!} =\frac{3}{7}[/m]
[m]\frac{x-y}{y+1}=\frac{3}{7}[/m]
[m]С^{y}_{x}:С^{y-1}_{x}=14:21[/m] ⇒ [m] \frac{x!}{(y)!\cdot (x-y)!}: \frac{x!}{(y-1)!\cdot (x-y+1)!}=\frac{2}{3}[/m]
[m] \frac{x!}{(y-1)!\cdot y\cdot (x-y)!}\cdot \frac{(y-1)!\cdot (x-y)!\cdot (x-y+1)}{x!}=\frac{2}{3}[/m]
[m]\frac{x-y+1}{y}=\frac{2}{3}[/m]
Система:
[m]\left\{\begin{matrix}
\frac{x-y}{y+1}=\frac{3}{7}\\ \frac{x-y+1}{y}=\frac{2}{3}\end{matrix}\right.[/m][m]\left\{\begin{matrix}
7x-7y=3y+3 \\ 3x-3y+3=2y\end{matrix}\right.[/m][m]\left\{\begin{matrix}
7x=10y+8\\ 3x=5y-3\end{matrix}\right.[/m][m]\left\{\begin{matrix}
x=\frac{10y+8}{7}\\ x=\frac{5y-3}{3}\end{matrix}\right.[/m]
Приравниваем правые части
[m]\frac{10y+8}{7}=\frac{5y-3}{3}[/m]
[m]30y+24=35y-21[/m]
[m]5y=45[/m] ⇒ [m]y=9[/m]
[m]x=\frac{10\cdot 9+8}{7}=14[/m]
О т в е т. [m] x=14[/m]; [m] y=9[/m];