Can we show $\binom{n}{r}=\binom{n-1}{r-1}+\binom{n-2}{r-1}+\binom{n-3}{r-1}+\dotsb+\binom{r-1}{r-1}$?

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Prove that \[\sum_{r=0}^{n}\binom{n}{r}=2^n.\]

By considering the number of ways of choosing a set of \(n\) people from a set of \(2n\) people, or otherwise, prove that \[\binom{2n}{n}=\binom{n}{0}^2+\binom{n}{1}^2+\dotsb+\binom{n}{n}^2.\]

Prove that \[\binom{n}{r}=\binom{n-1}{r-1}+\binom{n-1}{r}\] and hence, or otherwise, prove that \[\binom{n}{r}=\binom{n-1}{r-1}+\binom{n-2}{r-1}+\binom{n-3}{r-1}+\dotsb+\binom{r-1}{r-1},\] where \(r\), \(n\) are positive integers with \(1 \leq r \leq n-1\).