# How do you find three consecutive binomial coefficients in the relationship 1:2:3?

## They are $\left(\begin{matrix}14 \\ 4\end{matrix}\right)$, $\left(\begin{matrix}14 \\ 5\end{matrix}\right)$, and $\left(\begin{matrix}14 \\ 6\end{matrix}\right)$, but I'd like to know how to obtain that result without using Pascal's Triangle.

Nov 22, 2016

We can represent the desired binomial coefficients as

((n),(k)) = (n!)/(k!(n-k)!)
((n),(k+1))=(n!)/((k+1)!(n-k-1)!)
((n),(k+2))=(n!)/((k+2)!(n-k-2)!)

Then, based on the given ratios, we have

(((n),(k+1)))/(((n),(k)))= (k!(n-k)!)/((k+1)!(n-k-1)!) = (n-k)/(k+1) = 2

$\implies n - k = 2 k + 2$

$\implies n - 3 k = 2$

and

(((n),(k+2)))/(((n),(k+1)))=((k+1)!(n-k-1)!)/((k+2)!(n-k-2)!) = (n-k-1)/(k+2) = 3/2

$\implies 2 n - 2 k - 2 = 3 k + 6$

$\implies 2 n - 5 k = 8$

This gives us the system of equations

$\left\{\begin{matrix}n - 3 k = 2 \\ 2 n - 5 k = 8\end{matrix}\right.$

Solving this, we arrive at

$\left\{\begin{matrix}n = 14 \\ k = 4\end{matrix}\right.$

Thus, our binomial coefficients are

$\left(\begin{matrix}14 \\ 4\end{matrix}\right) , \left(\begin{matrix}14 \\ 5\end{matrix}\right) , \left(\begin{matrix}14 \\ 6\end{matrix}\right)$

as expected.