Pascal's rule

In the $n^{t h}$ row of Pascal's triangle, the $k^{t h}$ entry is denoted $(\binom{n}{k})$ , a binomial coefficient. By definition, when $k$ not in $0 \leq k \leq n$ , then $(\binom{n}{k}) = 0$ .

Each entry is the sum of the two previous entries.

We write:

(\binom{n}{k}) = (\binom{n - 1}{k - 1}) + (\binom{n - 1}{k})

(\binom{n + 1}{k}) = (\binom{n}{k - 1}) + (\binom{n}{k})

Visually,

Proof

\begin{aligned} (\binom{n - 1}{k}) + (\binom{n - 1}{k - 1}) & = \frac{(n - 1)!}{k! (n - 1 - k)!} + \frac{(n - 1)!}{(k - 1)! (n - k)!} \\ = (n - 1)! (\frac{n - k}{k! (n - k)!} + \frac{k}{k! (n - k)!}) \\ = (n - 1)! \frac{n}{k! (n - k)!} \\ = \frac{n!}{k! (n - k)!} \\ = (\binom{n}{k}) \end{aligned}

QED