Binomial coefficient formula reduction

Question

I think I need to use $(x+y)^n = \sum_{k=0}^n {n \choose k} x^{n-k} y^k$, but I can't figure out how. I am trying to see why

$$\frac{y!}{x!(y-x)!} \lambda^x \mu^{y-x} / \sum_{x,r:x+r=y}\frac{y!}{x! r!} \lambda^x \mu^r$$

is the same as

$${y \choose x} \left(\frac{\lambda}{\lambda + \mu} \right)^x \left( \frac{\mu}{\lambda + \mu} \right)^{y-x}$$

But can't see why

Answer 1

The numerator of your LHS is by definition of ${n \choose k}$ equal to $${y \choose x} \lambda^x \mu^{y-x}$$ The denominator can be rewritten as: $$\sum_{r=0}^y {y \choose r} \mu^r \lambda^{y-r} = (\mu+\lambda)^y = (\mu+\lambda)^x (\mu+\lambda)^{y-x}$$

Now if you take a ratio of the simplified numerator and denominator, you will get the desired answer,

Answer 2

If you match the two expressions and cancel terms, you find that you need to show $$(\lambda + \mu)^y = \sum_{x,r : x+r = y} \frac{y!}{x! r!} \lambda^x \mu^r.$$

Answer 3

It seems that the index r starts at $r=0$ and the maximum value of r is if $x=0$. That means $r=y-x=y-0=y$.

$$\sum_{r=0}^y\frac{y!}{x! r!} \lambda^x \mu^r \overset{substitution}{=}\sum_{r=0}^y\frac{y!}{(y-r)! r!}\lambda^{y-r} \mu^r$$

Now apply the binomial theorem.