I'm trying to prove Pascal's Identity algebraically but I'm getting stuck... I'm ashamed to say I spent an hour trying to do this with no luck. The various solutions I've seen seem to do steps that I don't understand. Would appreciate if someone could do a detailed walkthrough of this.
Here's what I have so far:
$\frac{(n−1)!}{k!(n−1−k)!} + \frac{(n−1)!}{(k−1)!(n−1−(k−1))!}$
$= \frac{(n−1)!}{k!(n−1−k)!} + \frac{(n−1)!}{(k−1)!(n−k)!}$
$= \frac{((n−1)!(k−1)!(n−k)! + (n−1)!k!(n−1−k)!}{k!(n−1−k)!(k−1)!(n−k)!}$ (I don't even know if this part is right... sigh)
Edit: Yes, there are other solutions, but I do not understand them. I'm looking for a more detailed step-by-step solution. I'm clearly having trouble with even basic algebra so sticking with that would be appreciated.
To continue with your approach:
\begin{align*} \frac{(n-1)!}{k!(n-1-k)!} + \frac{(n-1)!}{(k-1)!(n-1-(k-1))!} &= \frac{(n-1)!}{k!(n-1-k)!} + \frac{(n-1)!}{(k-1)!(n-k)!}\\ &\overset{(1)}= \frac{(n-k)\cdot(n-1)!}{k!(n-k)!} + \frac{k\cdot(n-1)!}{k!(n-k)!}\\ &= \frac{(n-k)\cdot(n-1)!+k\cdot(n-1)!}{k!(n-k)!} \\ &= \frac{(n-k+k)\cdot(n-1)!}{k!(n-k)!} \\ &= \frac{n\cdot(n-1)!}{k!(n-k)!} \\ &= \frac{n!}{k!(n-k)!} \end{align*}
$(1)$: Common denominator of $k!(n-1-k)!$ and $(k-1)!(n-k)!$ is $k!(n-k)!$