Question

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

They are `((14),(4))`, `((14),(5))`, and `((14),(6))`, but I'd like to know how to obtain that result without using Pascal's Triangle.

Answer 1

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`

`=>n-k = 2k+2`

`=> n-3k = 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`

`=> 2n-2k-2=3k+6`

`=> 2n-5k=8`

This gives us the system of equations

`{(n-3k = 2),(2n-5k=8):}`

Solving this, we arrive at

`{(n = 14), (k=4):}`

Thus, our binomial coefficients are

`((14),(4)), ((14),(5)), ((14),(6))`

as expected.