# Proving Pascal's Identity algebraically, stuck with simplification [duplicate]

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.

• Bring the two terms to a common denominator of $k!(n-k)!$ by multiplying the top and bottom of the first term by $n-k$, and the top and bottom of the second term by $k$. Then add the new tops. These will have sum $(n-1)!(n-k)+(n-1)!k$, which simplifies to $n!$. Aug 6, 2016 at 1:42
• Also, see under Related, right side of this page. You will find your problem, with solutions. But maybe first try to carry out the details of the above calculation yourself. Aug 6, 2016 at 1:46
• @AndréNicolas I saw your answer in another post, but I didn't understand it. The denominator of the first term is now $k!(n−1−k)!(n−k)$ and the second term is $(k-1)!(n-k)!k$, and they don't look the same to me. Aug 6, 2016 at 1:50
• Note that $(n-k-1)!(n-k)=(n-k)!$ and $(k-1)!k=k!$. Now they look the same. Aug 6, 2016 at 2:04
• Ohh. Thanks for the shortcuts. Can you also go through the process with just basic algebra? (e.g. multiply first term by $(k-1)!(n-k)!$.) Aug 6, 2016 at 2:15

$(1)$: Common denominator of $k!(n-1-k)!$ and $(k-1)!(n-k)!$ is $k!(n-k)!$