1. ## joint pmf

if X and Y are discrete RVs with pmf $\displaystyle p(x,y) = a\; \dfrac{k!}{x!y!(k-x-y)!}$ where x and y are$\displaystyle non \; negative \; integers \; and\; x+y\leq k$.

I want to find the value of a.

I know that I need to show that the pmf sums up to 1.

$\displaystyle \sum_x \sum_y p(x,y)=1$
since x and y are non negatve: 0<=x+y<=k
here i am getting confused in where x and y range from.
can anyone help me how to solve this?

2. $\displaystyle \sum_{x=0}^{k} \sum_{y=0}^{k-x} p(x,y)=1$

so,
$\displaystyle a \cdot \sum_{x=0}^{k} \sum_{y=0}^{k-x} \dfrac{k!}{x!y!(k-x-y)!} = a \cdot \sum_{x=0}^{k} \dfrac{k!\cdot 2^{k-x}}{x!(k-x)!}=a\cdot 3^k =1$

Hence $\displaystyle a=\left(\frac{1}{3}\right)^k$

3. Thank you so much.

But how do we get $\displaystyle \sum_{y=o}^{k-x}\dfrac{1}{y!(k-x-y)!}= \dfrac{2^{k-x}}{(k-x)!}$

is this a gamma function? Could you explain please.

4. Refer to the section on series of binomial coefficients
Binomial coefficient - Wikipedia, the free encyclopedia

$\displaystyle a \cdot \sum_{x=0}^{k} \sum_{y=0}^{k-x} \dfrac{k!}{x!y!(k-x-y)!} = a \cdot \sum_{x=0}^{k} \sum_{y=0}^{k-x} \dfrac{k!\cdot \left(k-x\right)!}{x!y!(k-x-y)!(k-x)!} =a \cdot \sum_{x=0}^{k} \dfrac{k!}{x!(k-x)!}\sum_{y=0}^{k-x} \dfrac{\left(k-x\right)!}{y!(k-x-y)!}$

$\displaystyle (Y+X)^n = \sum_{k=0}^n {n \choose k} X^{n-k}Y^K$
$\displaystyle a \cdot \sum_{x=0}^{k} \dfrac{k!\cdot 2^{k-x}}{x!(k-x)!}= a \cdot \sum_{x=0}^\infty {k \choose x} 2^{k-x}1^k=a \cdot \left(2+1\right)^k$