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$\dfrac{\dfrac{n!}{k!(n-k)!}}{\dfrac{n!}{(k+1)!(n-k-1)!}} = \\ \dfrac{(k+1)!(n-k-1)!}{k!(n-k)!}=\\ \dfrac{k+1}{n-k}=\dfrac{4}{11}$

$11k+11=4n-4k\\ 4n-15k=11$

$\text{We can use the Euclidean algorithm to find }n=14,~k=3$

Let

$n$

and

$k$

be positive integers such that $\dbinom{n}{k}:\dbinom{n}{k + 1} = 4:11$. Find the smallest possible value of .

$\begin{array}{|rcll|} \hline \dfrac{\dbinom{n}{k}} {\dbinom{n}{k + 1}} &=& \dfrac{4}{11} \quad | \quad \dbinom{n}{k + 1} = \dbinom{n}{k}\cdot \dfrac{n-k}{k+1} \\\\ \dfrac{\dbinom{n}{k}} {\dbinom{n}{k}\cdot \dfrac{n-k}{k+1}} &=& \dfrac{4}{11} \\\\ \dfrac{k+1}{n-k} &=& \dfrac{4}{11} \\\\ 11(k+1) &=& 4(n-k) \\ 11k+11 &=& 4n-4k \\ \mathbf{4n-15k} & \mathbf{=} & \mathbf{11} \\\\ 4n &=& 11+15k \\ n &=& \dfrac{11+15k} {4} \\ &=& \dfrac{11+1-1+15k+k-k} {4} \\ &=& \dfrac{12+16k-(1+k)} {4} \\ &=& \dfrac{12+16k} {4}- \dfrac{(1+k)} {4} \\ &=& 3+4k - \underbrace{\dfrac{(1+k)} {4}}_{=a} \\ \mathbf{n} &\mathbf{=}& \mathbf{3+4k - a} \\\\ a &=& \dfrac{1+k} {4} \\ 4a &=& 1+k \\ \mathbf{k} &\mathbf{=} & \mathbf{4a-1,\quad a\in \mathbb{Z}}\\\\ n &=& 3+4k - a \quad | \quad k=4a-1 \\ &=& 3+4(4a-1) - a \\ &=& 3+16a-4 - a \\ \mathbf{n} &\mathbf{=}& \mathbf{-1+15a} \\\\ n_{min} &=& -1+15\cdot 1\quad | \quad a=1 \\ \mathbf{n_{min}} &\mathbf{=}& \mathbf{14} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline k&=& 4a-1 \quad | \quad a=1 \\ k&=& 4\cdot1 -1 \\ k&=& 3 \\ \dfrac{\dbinom{14}{3}} {\dbinom{14}{4}} &=& \dfrac{4}{11} \\ \hline \end{array}$