I have to prove for x real and p integer that $\begin{pmatrix}x\\ p\end{pmatrix}=\left(-1\right)^p\:\cdot \begin{pmatrix}p-x-1\\ \:p\end{pmatrix}$

I have tried to remember all the properties of falling and rising factorials before starting working, i didn't find a relation of $\left(-1\right)^p\:\cdot \begin{pmatrix}p-x-1\\ \:p\end{pmatrix}$. So what i did was to use the definition of combinatorics.

  1. $\frac{x!}{p!\left(x-p\right)!}=\left(-1\right)^p\:\cdot \:\left(\frac{p-x-1}{p!\left(-x-1\right)!}\right)$

  2. $\frac{x!}{\frac{p!\left(x-p\right)!}{\frac{\left(p-x-1\right)!}{p!\left(-x-1\right)!}}}=\left(-1\right)^p$

  3. $\frac{\left(-x-1\right)!x!}{\left(x-p\right)!\left(p-x-1\right)!}=\left(-1\right)^p$

  4. $x!\left(-x-1\right)!=\left(-1\right)^p\:\cdot \:\left(x-p\right)!\left(p-x-1\right)!$


This is an exercise in the definition of generalized binomial coefficients, use of product notation, and manipulation of indices.

$\begin{array}\\ \binom{x}{p} &=\dfrac{\prod_{k=0}^{p-1}(x-k)}{p!} \qquad\text{(Definition of generalized binomial coefficient)}\\ &=\dfrac{(-1)^p\prod_{k=0}^{p-1}(k-x)}{p!} \qquad\text{(Reverse the sign of each term)}\\ &=(-1)^p\dfrac{\prod_{k=0}^{p-1}((p-1-k)-x)}{p!} \qquad\text{(Reverse the order of the product so } k \to p-1-k)\\ &=(-1)^p\dfrac{\prod_{k=0}^{p-1}(p-x-1-k)}{p!} \qquad\text{(Rearrange the expression)}\\ &=(-1)^p \binom{p-x-1}{p} \qquad\text{(Definition, again)}\\ \end{array} $


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