Having $\binom{\lambda}{n}=n+1$, which is $\lambda$?

Question

One question, if I've got this, $\binom{\lambda}{n}=n+1$, which has to be the value of $\lambda$?

Answer 1

Let $\lambda$ and $n$ be natural numbers such that $\lambda \ge n$.

If $n=0$ then $\lambda$ is any natural number in fact $\binom{\lambda}{0}=1$ for all $\lambda \in \mathbb{N}$.

Now we suppose that $n \ge 1$.

If $\lambda>n+1$ then $\binom{\lambda}{n}=\frac{\lambda(\lambda-1)(\lambda-2)\cdot...\cdot(\lambda-n+1)}{n!}>\frac{(n+1)n(n-1)\cdot...\cdot2}{n!}=\frac{(n+1)!}{n!}=n+1$.

Moreover, if $\lambda=n$ then $\binom{\lambda}{n}=1<n+1$.

Therefore, if $n\ge1$, there is just one possibility that is $\lambda=n+1$.

$\binom{\lambda}{n}=\binom{n+1}{n}=n+1$.

So in any case:

if $n=0$, $\lambda$ is any natural number

if $n\ge1$, $\lambda=n+1$

Answer 2

Hint: recall that $$x! = \Gamma(x+1)$$ Now solve the equation using Euler gamma function, and then restrict the solutions to $\lambda \in \mathbb{N}$.