Binomial Coefficient with Self minus One

Theorem

$\forall n \in \N_{>0}: \dbinom n {n - 1} = n$

The case where $n = 1$ can be taken separately.

$\dbinom 1 0 = 1$

demonstrating that the result holds for $n = 1$.

Let $n \in \N: n > 1$.

$\dbinom n {n - 1} = \dfrac {n!} {\paren {n - 1}! \paren {n - \paren {n - 1} }!} = \dfrac {n!} {\paren {n - 1}! \ 1!}$

the result following directly from the definition of the factorial.

$\blacksquare$

From Cardinality of Set of Subsets, $\dbinom n {n - 1}$ is the number of combination of things taken $n - 1$ at a time.

Choosing $n - 1$ things from $n$ is the same thing as choosing which $1$ of the elements to be left out.

There are $n$ different choices for that $1$ element.

Therefore there are $n$ ways to choose $n - 1$ things from $n$.

$\blacksquare$