One Choose n

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Theorem

$\dbinom 1 n = \begin{cases} 1 & : n \in \left\{ {0, 1}\right\} \\ 0 & : \text {otherwise} \end{cases}$

where $\dbinom 1 n$ denotes a binomial coefficient.


Proof

By definition of binomial coefficient:

$\dbinom m n = \begin{cases}

\dfrac {m!} {n! \left({m - n}\right)!} & : 0 \le n \le m \\ & \\ 0 & : \text { otherwise } \end{cases}$

Thus when $n > 1$:

$\dbinom 1 n = 0$

and when $n < 0$:

$\dbinom 1 n = 0$

Then:

\(\ds \dbinom 1 0\) \(=\) \(\ds \dfrac {1!} {0! \left({0 - 1}\right)!}\) Definition of Binomial Coefficient
\(\ds \) \(=\) \(\ds 1\) Definition of Factorial


\(\ds \dbinom 1 1\) \(=\) \(\ds \dbinom 1 {1 - 1}\) Symmetry Rule for Binomial Coefficients
\(\ds \) \(=\) \(\ds \dbinom 1 0\)
\(\ds \) \(=\) \(\ds 1\) from above

$\blacksquare$