Given that $^_ = \frac{3}{8} \times{}^_{+1}$ and $^_ ={}^_{+6}$, find and .

Question

Given that $^_ = \frac{3}{8} \times{}^_{+1}$ and $^_ ={}^_{+6}$, find $$ and $$.

Can anyone see a slick way to do this? The solution is $x=10$ and $y=2$, but I am struggling with the method.

Answer 1

$$\overbrace{\frac 38 \binom x{y+1}=\binom xy}^\text{given}=\frac {y+1}{x-y}\binom x{y+1}\\ \frac {y+1}{x-y}=\frac 38\\ 3x=11y+8\qquad\cdots (1)\\\\ \binom x{\color{orange}{y}}=\binom x{\color{blue}{y+6}}\\ \binom x{\color{blue}{x-y}}=\binom x{\color{orange}{x-y-6}}\\ x-y=y+6\\ x=2y+6\qquad\cdots (2)\\ \text {From }(1),(2), \hspace{6cm}\\ \color{red}{x=10, y=2}$$

Answer 2

Hint: The first relation gives \begin{align}\dbinom{x}{y}=\frac38\dbinom{x}{y+1}\implies& \frac{x!}{y!(x-y)!}=\frac38\frac{x!}{(y+1)!(x-y-1)!}\\[0.3cm]\implies&\frac{x!}{y!(x-y)(x-y-1)!}=\frac38\frac{x!}{(y+1)y!(x-y-1)!}\\[0.3cm]\implies&\frac{1}{(x-y)}=\frac38\frac{1}{(y+1)}\implies -3x+11y=-8\end{align} For the second, recall that $$\dbinom{x}{y}=\dbinom{x}{x-y}$$ which gives that $x-y=y+6$.