Symmetry Rule for Binomial Coefficients/Proof 1
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Theorem
Let $n \in \Z_{>0}, k \in \Z$.
Then:
- $\dbinom n k = \dbinom n {n - k}$
Proof
Follows directly from the definition of binomial coefficient, as follows.
If $k < 0$ then $n - k > n$.
Similarly, if $k > n$, then $n - k < 0$.
In both cases:
- $\dbinom n k = \dbinom n {n - k} = 0$
Let $0 \le k \le n$.
\(\ds \binom n k\) | \(=\) | \(\ds \frac {n!} {k! \paren {n - k}!}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {n!} {\paren {n - k}! k!}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {n!} {\paren {n - k}! \paren {n - \paren {n - k} } !}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \binom n {n - k}\) |
$\blacksquare$
Sources
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text I$. Algebra: Permutations and Combinations: Two important relations
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.6$: Binomial Coefficients: $\text{B}$