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I need to prove ${n \choose k} = {n \choose k -1}\frac{n -k +1}{k}$ by induction for k in {2 .....,n}. Can someone show be the proof? Here is the work I have done for the inductive step: Assume true for ${n \choose k} = {n \choose k -1}\frac{n -k +1}{k}$ We need to show ${n+1 \choose k} = {n+1 \choose k -1}\frac{n+1 -k +1}{k}$

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

${n \choose k} + {n \choose k-1} = ({n \choose k -1}+{n \choose k -2})\frac{n+2 -k}{k}$

Then by the inductive hypothesis:

${n \choose k -1}\frac{n -k +1}{k} +{n \choose k -2}\frac{n -k +2}{k-1} = ({n \choose k -1}+{n \choose k -2})\frac{n+2 -k}{k}$

I cannot figure out where to go from here

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    How much of it can you do? What have you tried? – 2017-02-14
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    No, we are not going to do *all* your work for you; it should be your own work. We will *help out* if you show us what you have tried and where you get stuck, so do please show your efforts. – 2017-02-14

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