I am solving a probability question from my math course, and I have gotten to the following expression:
$$\left(\sum_{i=1}^n \frac{(x+y+i-2)!}{y ! (x-1)! (i-1)!}\right) \frac{n! x! y!}{(n + x + y)!}$$
However, the final solution is such:
$$\frac{n x}{(x + y) (x + y + n) }$$
Using Wolfram Mathematica, I confirmed that my equation and the final answer are equivalent, but how would I go about simplifying my answer to the get final answer by hand?
In case anyone is curious, the question asks that if there are $x$ jelly donuts, $y$ chocolate donuts, and $n$ vanilla donuts in a box, what is the probability that if they are taken out and subsequently eaten one by one uniformly at random, that the chocolate donuts will be the first type of donut to be completely finished, the jelly donuts will be the second type to be finished, and the vanilla donuts will be the last to be finished?
Any alternate solutions are also appreciated, thanks for the help.
About the algebraic simplification, first note that :
$$\sum_{i=1}^{n}\frac{(x+y+i-2)!}{y!(x-1)!(i-1)!}\times \frac{n!x!y!}{(n+x+y)!}=\sum_{i=0}^{n-1}\frac{(x+y+i-1)!}{i!}\times \frac{x\times n!}{(n+x+y)!}$$
Here, I have just extracted the terms that do not depend on $i$ from the sum, and made the change of variable "$i=i-1$". Then you can make appear a coefficient binomial in the sum by dividing by $(x+y-1)!$ to get :
$$\sum_{i=0}^{n-1}\frac{(x+y+i-1)!}{i!}=\sum_{i=0}^{n-1}\binom{x+y+i-1}{i}\times (x+y-1)!$$
Then apply the formula
$$\sum_{i=0}^{n-1} \binom{X+i}{i}=\binom{X+n}{n-1}$$
(you can find a proof of this formula here : Prove that $C(n,0) + C(n+1,1) +\dots+ C(n+r,r) = C(n+r+1,r).$ )
where $X=x+y-1$ to get :
$$\sum_{i=0}^{n-1}\frac{(x+y+i-1)!}{i!}\times \frac{x\times n!}{(n+x+y)!} =\binom{n+x+y-1}{n-1}\times \frac{x\times (x+y-1)!n!}{(n+x+y)!}$$
Simplify and you get the result.