We know that the gamma function is the following $\Gamma(x)=(x-1)!= (x-1)\times (x-2)\times...\times1$
Suppose that we have the following expression
$$\prod_{i=0}^{n-1}(x+i)^{I(y_{i+1}\in A)}=(x+0)^{I(y_{1}\in A)}\times (x+1)^{I(y_{2}\in A)}\times...\times(x+n-2)^{I(y_{n-1}\in A)}\times(x+n-1)^{I(y_{n}\in A)}$$
If all the indicator functions were equal to $1$, i.e $I(y_{i}\in A)=1$ and $y_{i}$ doesn't have actual meaning is just to point out which terms are equal to $1$ or not, then the previous expression could be written as
$$\frac{\Gamma(x+n)}{\Gamma(x)}=\frac{1\times...(x-1)\times(x+0)\times (x+1)\times...\times(x+n-2)\times(x+n-1)}{1\times...\times(x-1)}$$
$$=(x+0)\times (x+1)\times...\times(x+n-2)\times(x+n-1)$$
So, my question is:
How, do we modify the gamma function in order to account for the indicator functions and still derive something of the form $\frac{\Gamma(x+n)}{\Gamma(x)}$.