Binomial Coefficients Proof Help

Question

I'm having trouble doing these proofs. So far I've changed the N choose R into its factorial form and have simplified but I'm stuck.

$\left( \begin{array} { c } { n } \\ { r } \end{array} \right) = \frac { n - r + 1 } { r } \cdot \left( \begin{array} { c } { n } \\ { r - 1 } \end{array} \right)$

$\left( \begin{array} { l } { n } \\ { r } \end{array} \right) = \frac { n } { n - r } \cdot \left( \begin{array} { c } { n - 1 } \\ { r } \end{array} \right)$

$n \left( \begin{array} { c } { n - 1 } \\ { r } \end{array} \right) = ( r + 1 ) \left( \begin{array} { c } { n } \\ { r + 1 } \end{array} \right)$

Answer 1

$\frac{n-r+1}{r}\binom{n}{r-1}=\frac{n-r+1}{r}\frac{n!}{(r-1)!(n-r+1)!}=\frac{n!}{r(r-1)!(n-r)!}=\frac{n!}{r!(n-r)!}=\binom{n}{r}$

That was the first identity. Can you do the other two now? Exactly the same idea.

Answer 2

You have the well-know formula $${n\choose k} = \frac{n!}{k!(n-k)!}$$

So, for the first equality : $${n\choose r} = \frac{n!}{r!(n-r)!} = \frac{n! \times (n-r+1)}{(r-1)! \times r \times (n-r+1) !} = \frac{n-r+1}{r} {n \choose r-1}$$

You can do the two others in the exact same way !