# 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.

Mar 16, 2018

The ratio of two consecutive binomial coefficients is given by :

(((n),(r+1)))/(((n),(r)))= (n!)/((r+1)!(n-r-1)!) times (r!(n-r)!)/(n!) = (n-r)/(r+1)

So, for $\left(\begin{matrix}n \\ r\end{matrix}\right) , \left(\begin{matrix}n \\ r + 1\end{matrix}\right)$, and $\left(\begin{matrix}n \\ r + 2\end{matrix}\right)$ to be in the ratio 1:2:3, we must have

$\frac{n - r}{r + 1} = 2 \implies n = 3 r + 2$

and

$\frac{n - r - 1}{r + 2} = \frac{3}{2} \implies n = \frac{5}{2} r + 4$

The two relations together give

$3 r + 2 = \frac{5}{2} r + 4 \implies \frac{r}{2} = 2 \implies r = 4$.

Using either of the two relations then leads to $n = 14$