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 |
$\blacksquare$