5.034 (а). Найти x и y, если:
$$С^y_{x+1} : C^{y+1}_x : C^{y-1}_x=6:5:2.$$
Решение.
\[C^y_{x+1}:C^{y+1}_x:C^{y-1}_x=6:5:2\]
\(\left\{\begin{array}{lcl}C^y_{x+1}:C^{y+1}_x=6:5\\C^{y+1}_x:C^{y-1}_x=5:2\end{array}\right.\)
\(C_n^m = \frac{n!}{(n - m)! m!}\), где \(m \leq n; C_n^0 = 1;\)
\(C^y_{x+1}=\frac{(x+1)!}{(x+1-y)! \cdot y!}=\frac{(x+1) \cdot x!}{(x-y+1)(x-y)(x-y-1)! \cdot y!}\)
\(C^{y+1}_x=\frac{x!}{(x-(y+1))! \cdot (y+1)!}=\frac{x!}{(x-y-1)! \cdot (y+1)!}=\frac{x!}{(x-y-1)! \cdot (y+1) \cdot y!}\)
\(C^{y-1}_x=\frac{x!}{(x-(y-1))! \cdot (y-1)!}=\frac{x!}{(x-y+1)! \cdot (y-1)!}=\frac{x!}{(x-y+1)(x-y)(x-y-1)! \cdot (y-1)!}\)
\(\left\{\begin{array}{lcl}\frac{(x+1) \cdot x!}{(x-y+1)(x-y)(x-y-1)! \cdot y!}:\frac{x!}{(x-y-1)! \cdot (y+1) \cdot y!}=6:5  \\ \frac{x!}{(x-y-1)! \cdot (y+1) \cdot y!}:\frac{x!}{(x-y+1)(x-y)(x-y-1)! \cdot (y-1)!}=5:2 \end{array}\right.\)
\(\left\{\begin{array}{lcl}\frac{(x+1) \cdot x!}{(x-y+1)(x-y)(x-y-1)! \cdot y!}\cdot\frac{(x-y-1)! \cdot (y+1) \cdot y!}{x!}=\frac{6}{5} \\ \frac{x!}{(x-y-1)! \cdot (y+1) \cdot y!}\cdot\frac{(x-y+1)(x-y)(x-y-1)! \cdot (y-1)!}{x!}=\frac{5}{2} \end{array}\right.\)
\(\left\{\begin{array}{lcl}\frac{(x+1)}{(x-y+1)(x-y)}\cdot\frac{(y+1)}{1}=\frac{6}{5} \\ \frac{1}{(y+1) \cdot y \cdot (y-1)!}\cdot\frac{(x-y+1)(x-y) \cdot (y-1)!}{1}=\frac{5}{2} \end{array}\right.\)
\(\left\{\begin{array}{lcl}\frac{(x+1)}{(x-y+1)(x-y)}\cdot\frac{(y+1)}{1}=\frac{6}{5}  \\ \frac{1}{(y+1) \cdot y}\cdot\frac{(x-y+1)(x-y)}{1}=\frac{5}{2} \end{array}\right.\)
\(\left\{\begin{array}{lcl}\frac{(x+1)(y+1)}{(x-y+1)(x-y)}=\frac{6}{5}  \\ \frac{(x-y+1)(x-y)}{(y+1) \cdot y}=\frac{5}{2} \end{array}\right.\)
\(\left\{\begin{array}{lcl}\frac{(x-y+1)(x-y)}{(x+1)(y+1)}=\frac{5}{6} \\ \frac{(x-y+1)(x-y)}{(y+1) \cdot y}=\frac{5}{2} \end{array}\right.\)
\(\frac{(x-y+1)(x-y)}{(y+1) \cdot y}=\frac{5}{2}\)
\(\frac{(x-y+1)(x-y)}{y}\cdot\frac{1}{y+1}=\frac{5}{2}\)
\(\frac{(x-y+1)(x-y)}{y}=\frac{5}{2}:\frac{1}{y+1}\)
\(\frac{(x-y+1)(x-y)}{y}=\frac{5}{2}\cdot(y+1)\)
\(\frac{(x-y+1)(x-y)}{y}:\frac{5}{2}=y+1\)
\(y+1=\frac{(x-y+1)(x-y)}{y}\cdot\frac{2}{5}\)
\(\left\{\begin{array}{lcl}\frac{(x-y+1)(x-y)}{(x+1)(y+1)}=\frac{5}{6} \\ y+1=\frac{(x-y+1)(x-y)}{y}\cdot\frac{2}{5} \end{array}\right.\)
Подставляем вместо \(y+1\) полученное значение:
\(\frac{(x-y+1)(x-y)}{(x+1)\cdot\frac{(x-y+1)(x-y)}{y}\cdot\frac{2}{5}}=\frac{5}{6}\)
\(\frac{(x-y+1)(x-y)}{\frac{2(x+1)(x-y+1)(x-y)}{5y}}=\frac{5}{6}\)
\((x-y+1)(x-y):\frac{2(x+1)(x-y+1)(x-y)}{5y}=\frac{5}{6}\)
\((x-y+1)(x-y)\cdot\frac{5y}{2(x+1)(x-y+1)(x-y)}=\frac{5}{6}\)
\(\frac{y}{(x+1)}\cdot\frac{5}{2}=\frac{5}{6}\)
\(\frac{y}{(x+1)}=\frac{5}{6}:\frac{5}{2}\)
\(\frac{y}{(x+1)}=\frac{5}{6}\cdot\frac{2}{5}\)
\(\frac{y}{(x+1)}=\frac{1}{3}\)
\(3y=x+1 \Rightarrow x+1=3y \Rightarrow x=3y-1\)
\(\left\{\begin{array}{lcl}x=3y-1 \\ y+1=\frac{(x-y+1)(x-y)}{y}\cdot\frac{2}{5} \end{array}\right.\)
\(y+1=\frac{(3y-1-y+1)(3y-1-y)}{y}\cdot\frac{2}{5}\)
\(y+1=\frac{2y\cdot(2y-1)}{y}\cdot\frac{2}{5}\)
\(y+1=\frac{2\cdot(2y-1)}{1}\cdot\frac{2}{5}\)
\(y+1=(4y-2)\cdot\frac{2}{5} | \cdot5\)
\(5\cdot(y+1)=(4y-2)\cdot2\)
\(5y+5=8y-4\)
\(5+4=8y-5y\)
\(3y=9\)
\(y=3\Rightarrow x=3y-1\Rightarrow x=3\cdot3-1=9-1=8\), то есть \(x=8.\)
Проверка.
\(C_{8+1}^3=C_9^3=\frac{9!}{(9-3)!\cdot3!}=\frac{9\cdot8\cdot7\cdot6!}{6!\cdot3\cdot2}=\frac{9\cdot8\cdot7}{3\cdot2}=\frac{3\cdot4\cdot7}{1}=84\)
\(C_8^{3+1}=C_8^4=\frac{8!}{(8-4)!\cdot4!}=\frac{8\cdot7\cdot6\cdot5\cdot4!}{4!\cdot4\cdot3\cdot2}=\frac{8\cdot7\cdot6\cdot5}{4\cdot6}=\frac{2\cdot7\cdot5}{1}=70\)
\(C_8^{3-1}=C_8^2=\frac{8!}{(8-2)!\cdot2!}=\frac{8\cdot7\cdot6!}{6!\cdot2}=\frac{4\cdot7}{1}=28\)
\(С_{x+1}^y:C^{y+1}_x:C^{y-1}_x=C_9^3:C_8^4:C_8^2=84:70:28=6:5:2\)
\(С_{x+1}^y:C^{y+1}_x:C^{y-1}_x=6:5:2\)
Значит, \(x=8\) и \(y=3\).
Ответ: \(x=8\); \(y=3\).

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