# Number of ways to fill a bag using red, blue and white balls

There are $$x$$ balls of color red, $$y$$ balls of color blue and $$z$$ balls of color white. The white balls can be individually painted to any color (either red or blue). And there is a bag, which can be filled with red and blue balls. In how many ways can the bag be filled in such a way that the number of red balls in the bag is same as the number of blue balls?

(Note that it is not necessary to use all the balls.)

I have tried the above problem, and I can get the solution by summation of possibilities of number of red/blue balls in the bag (like compute for case 1- 1R and 1B ball, case2- 2R and 2B balls, and so on and add them up). It would be really helpful if this equation can be simplified to a single formula (or maybe a recurrence relation that can be simplified?) without requiring any summation, so that the value can be calculated in constant time.

PS: I'm sorry I really couldn't find a better title.

If $$x\ge y$$, you must paint $$k$$ of the white balls red and $$k+d$$ of them blue, with $$d:+x-y$$, for some $$k$$ satisfying $$0\le k\le\lfloor\frac{z-d}{2}\rfloor$$. The number of solutions with indistinguishable white balls is then $$\sum_{k=0}^{\lfloor\frac{z-d}{2}\rfloor}\binom{z}{k}\binom{z-k}{k+d}$$. This formula generalises provided we define $$d:=|x-y|$$.
I consider the painted balls as indistinguishable. Denote by $$2M$$ the maximal realizable number of balls in the bag. The number of different ways to legally fill the bag then is $$M+1$$.
For the computation of $$M$$ assume $$x\leq y$$ with $$y-x=:d$$. Then we try to paint $$d$$ white balls red, and if white balls remain we paint them red and blue in equal numbers. This leads to M=\left\{\eqalign{&x+z\qquad\ \quad\qquad(z\leq d) \cr &y+\left\lfloor{z-d\over2}\right\rfloor\qquad(z\geq d)\ .\cr}\right.