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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.

userrandomnums
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    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 $n!$. – André Nicolas Aug 06 '16 at 01:42
  • 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. – André Nicolas Aug 06 '16 at 01:46
  • @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. – userrandomnums Aug 06 '16 at 01:50
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    Note that $(n-k-1)!(n-k)=(n-k)!$ and $(k-1)!k=k!$. Now they look the same. – André Nicolas Aug 06 '16 at 02:04
  • Ohh. Thanks for the shortcuts. Can you also go through the process with just basic algebra? (e.g. multiply first term by $(k-1)!(n-k)!$.) – userrandomnums Aug 06 '16 at 02:15
  • @AndréNicolas Okay I understand those equivalences you mentioned now. Makes much more sense now. Would still appreciate the algebra step-by-step if you have time :)! If not, no worries! – userrandomnums Aug 06 '16 at 03:40
  • I just used basic algebra, but used a "small" common denominator rather than a large one that would require a messy cancellation step at the end. – André Nicolas Aug 06 '16 at 04:53

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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 $k!(n-k)!$

Martin Sleziak
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