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I am trying to prove $$\binom{n}{k} = \binom{n}{k-1}\frac{n-k+1}{k}$$ for each $k \in \{1,...,n\}$ by induction. My professor gave us a hint for the inductive step to use the following four equations:
\begin{align*} \binom{n + 1}{k} & = \binom{n}{k} + \binom{n}{k - 1}\\ \binom{n + 1}{k - 1} & = \binom{n}{k - 1} + \binom{n}{k - 2}\\ \binom{n}{k} & = \binom{n}{k - 1}\frac{n - k + 1}{k}\\ \binom{n}{k - 1} & = \binom{n}{k - 2}\frac{n - k + 2}{k - 1} \end{align*} I keep getting stuck in the inductive step. I was hoping someone could help me.

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  • $\begingroup$ What's your definition of $\binom nk$? For instance, mine is $$\binom nk=\frac{n!}{k!(n-k)!}$$ which makes this theorem trivial (and the request of a proof by induction unreasonabe). $\endgroup$ – user228113 Mar 2 '16 at 20:56
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    $\begingroup$ The third equation is what you want to prove! So I assume you can only use it up to k? Why are the other equations acceptable? $\endgroup$ – fleablood Mar 2 '16 at 20:58
  • $\begingroup$ From the inductive hypothesis, I can assume the third equation is true. Thus making k $\epsilon$ {2,...,n+1}, we have k-1 $\epsilon$ {1,...,n} and then the inductive hypothesis becomes the fourth equation. The other equations are acceptable because they are by definition the recurrence relation for Pascal's triangle which has already been proved. $\endgroup$ – Tammy Mar 2 '16 at 21:01
  • $\begingroup$ Are you required to use induction? Simplifying the RHS is easier. $\endgroup$ – N. F. Taussig Mar 2 '16 at 21:01
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    $\begingroup$ Yes, I am required to prove by induction. I have already covered the case for when k=1 or k=n+1. Now I'm stuck with k $\epsilon$ {2,...,n}. Or finding n+1. So I'm pretty much just trying to use algebra with the above equations to find ${n+1 \choose k}$=${n+1 \choose k-1}$$\frac{n-k+2}{k}$ $\endgroup$ – Tammy Mar 2 '16 at 21:05
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Assume that $$\binom{n}{k} = \binom{n}{k-1}\frac{n-k+1}{k}$$

We need to prove:

$$\binom{n}{k+1} = \binom{n}{k}\frac{n-k}{k+1}$$

We will have

$$\begin{equation}\begin{aligned} \binom{n}{k+1} &= \frac{n!}{(k+1)!(n-k-1)!} \\ &= \frac{n!}{k!\times(k+1)\times1\times2\times...\times{(n-k)\div(n-k)}} \\ &= \frac{n!}{k!(n-k)!}\times\frac{n-k}{k+1} \\ &=\binom{n}{k-1}\times\frac{n-k+1}{k}\times\frac{n-k}{k+1}\\ &=\frac{n!}{(n-k+1)!(k-1)!}\times\frac{n-k+1}{k}\times\frac{n-k}{k+1}\\ &=\frac{n!(n-k+1)}{(n-k)!\times(n-k+1)\times(k-1)!\times k}\times\frac{n-k}{k+1}\\ &=\frac{n!}{k!(n-k)!}\times\frac{n-k}{k+1}\\ &=\binom{n}{k}\frac{n-k}{k+1} \end{aligned}\end{equation}$$

If $P(k)$ is true then $P(k+1)$ is also true, we can easily prove that $P(1)$ is true, so the induction is complete.

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