What are $n$ and $r$ in this question?

Question

This problem is puzzling me and I am not able to solve it on my own.

Let $$\binom nr=\binom n{r-1}$$ and $$\binom {n+1}r=\frac{40}{r!}\times \frac{n!}{(n-r+2)!}$$ What are $n$ and $r$ ?

Answer 1

$$\qquad \binom nr=\binom n{r-1}\implies 2r-1=n$$ $$\binom {n+1}r=\frac{40}{r!}\times \frac{n!}{(n-r+2)!}\implies n+1=\frac{40}{n-r+2}$$ thus $$ r=\frac{20}{r+1}\implies r=4$$ and $n=7$