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${n-1 \choose k-1} + {n-1 \choose k} = {n \choose k}$

My start: $$\begin{align}{n-1 \choose k-1} + {n-1 \choose k} &= \frac{(n-1)!}{(k-1)!(n-k)!} + \frac{(n-1)!}{(k!)(n-k-1)!}\\ &= (n-1)! \times \Big(\frac{1}{(k-1)!(n-k)!} + \frac{1}{(k)!(n-k-1)!}\Big)\\ &= (n-1)! \times \Big(\frac{1}{(k-1)!}\frac{1}{(n-k)!}+\frac{1}{k!}\frac{1}{(n-k-1)!}\Big)\end{align}$$

Now what do I do with what I have inside the parentheses?

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    the real equality is ${n-1 \choose k-1} + {n-1 \choose k} = {n \choose k}$ – 2017-01-29
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    Why did you change the question so drastically? You originally asked a specific question about numerators, and now you've revised the question to mean something else entirely. Fixing the times to plus is fine, rephrasing the question to ask something different makes me feel like I wasted effort trying to answer your original question. – 2017-01-29
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    (And your last line just repeats your second-to-last line.) – 2017-01-29

2 Answers 2

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The answer is that $$\frac{k}{k!}=\frac{k}{(k-1)!\cdot k}=\frac{1}{(k-1)!}$$

and similarly:

$$\frac{n-k}{(n-k)!}=\frac{1}{((n-1)-k)!}$$

They are just skipping over that step when factoring out $(n-1)!$:

$$\begin{align}\binom{n-1}{k-1}+\binom{n-1}{k}&=(n-1)!\left(\frac 1{(k-1)!(n-k)!}+\frac 1{k!(n-1-k)!}\right)\\ &=(n-1)!\left(\frac{k}{k!(n-k)!}+\frac{n-k}{k!(n-k)!}\right)\end{align}$$

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The real identity is:

$${n-1 \choose k-1}+ {n-1 \choose k} = {n \choose k}$$

and you can prove it like this:

$$\begin{align}{n-1 \choose k-1} + {n-1 \choose k} &= \frac{(n-1)!}{(k-1)!(n-k)!}+\frac{(n-1)!}{(k)!(n-k-1)!}\\&= \frac{(n-1)!}{(k-1)!(n-k-1)!}\left(\frac{1}{n-k}+\frac{1}{k}\right)\\ &=\frac{(n-1)!}{(k-1)!(n-k-1)!}\left(\frac{n}{(n-k)k}\right)\\& =\frac{n!}{k!(n-k)!}={n \choose k} \end{align}$$

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    @Thomas Andrews: thanks for the improvement. – 2017-01-29