If you want to proof $\binom{n}{k}$ + $\binom{n}{k+1}$ = $\binom{n+1}{k+1}$, you have to differentiate between 5 cases. Why 5 cases and where do they come from ? It should have something to do with $\binom{N}{k}$ = $\frac{n!}{k!(n-k)!}$ .
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"you have to differentiate between five cases" What? Says who? "Why 5 cases and where do they come from?" It sounds like you have a particular proof that you want explained. Share that proof. – JMoravitz Jul 22 '20 at 12:24
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This was a task at my college. You don't have to, but they want this 5 cases; k< -1, k = 1, k = -1 , k = n, k >n, 0<k<n. – Seenes Jul 22 '20 at 12:26
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At best guess... I suppose you may be referring to cases such as $k>n>0$, $k=n>0$, $k=n=0$, $k<-1$ and $n<0$ or something... but usually we don't bother with all of these at the time of this proof and just specify that $0\leq k<n$ as a hypothesis. As for why five cases and not more or less? That depends entirely on how you choose to organize yourself... – JMoravitz Jul 22 '20 at 12:27
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Exactly, but I don't get it why do I have to do this? What is the purpose of doing this? – Seenes Jul 22 '20 at 12:31
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Why should you restrict yourself to the case where $0\leq k<n$? Because outside of that range you'll have one or all of the terms in your identity equal to zero since they don't correspond to a valid counting question... Choosing $-3$ objects out of $7$ is impossible, so $\binom{7}{-3}$ is undefined or zero. Similarly, choosing ten objects out of five is impossible, so $\binom{5}{10}$ is undefined or zero, depending on how you prefer to define it (zero is more common). Why should you pay attention to any case other than the above? You shouldn't in my opinion. – JMoravitz Jul 22 '20 at 12:38
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Yes, thank u. I've understood the whole situation now. – Seenes Jul 22 '20 at 13:23