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Show that (n−1  r−1)  +( n−1 r ) = (n r)

this is combinations

I would apperatice your help :D

May 12, 2021

#1
+25988
+1

Show that $$\dbinom{n-1}{r-1} + \dbinom{n-1}{r} = \dbinom {n}{r}$$

$$\begin{array}{|rcll|} \hline && \mathbf{ \dbinom{n-1}{r-1} } \\ &=& \dfrac{(n-1)!} {(r-1)!\left(n-1-(r-1)\right)!} \\\\ &=& \dfrac{(n-1)!} {(r-1)!(n-1-r+1)!} \\ \\ &=& \dfrac{(n-1)!} {(r-1)!(n-r)!} \\ \\ && \boxed{ (n-1)! = \dfrac{n!}{n}\\\\(r-1)! = \dfrac{r!}{r} } \\\\ \mathbf{ \dbinom{n-1}{r-1} } &=& \mathbf{ \dfrac{n!*r} {n*r!(n-r)!} } \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline && \mathbf{ \dbinom{n-1}{r} } \\ &=& \dfrac{(n-1)!} {r!(n-1-r)!} \\\\ &=& \dfrac{(n-1)!} {r!(n-r-1)!} \\ \\ && \boxed{ (n-1)! = \dfrac{n!}{n}\\(n-r-1)! = \dfrac{(n-r)!}{n-r} } \\\\ \mathbf{ \dbinom{n-1}{r} }&=& \mathbf{ \dfrac{n!*(n-r)} {n*r!(n-r)!} } \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline && \dbinom{n-1}{r-1} + \dbinom{n-1}{r} \\\\ &=& \dfrac{n!*r} {n*r!(n-r)!} + \dfrac{n!*(n-r)} {n*r!(n-r)!} \\\\ &=& \dfrac{n!}{r!(n-r)!} \left( \dfrac{r+(n-r)}{n} \right) \\\\ &=& \dfrac{n!}{r!(n-r)!} \left( \dfrac{n}{n} \right) \\\\ &=& \dfrac{n!}{r!(n-r)!} \\\\ \dbinom{n-1}{r-1} + \dbinom{n-1}{r}&=& \mathbf{ \dbinom {n}{r} } \\ \hline \end{array}$$

May 13, 2021