Finding a distribution from the sum of two random variables in a joint trinomial distribution Let X and Y have a trinomial distribution with join probability function:
$ f(x,y) = \frac{n!}{x!y!(n-x-y)!}p^x q^y (1-p-q)^{n-x-y}; x=0,1,...n, y=0,1,...n$
Let $T=X+Y$
I can't seem to figure out where to go exactly. Here's where I am at:
$\sum_{x=0}^{n} \frac{n!}{x!y!(n-x-y)!}p^x q^y (1-p-q)^{n-x-y}\mbox{ substitute in Y=T-X}$
$\sum_{x=0}^{n} \frac{n!}{x!(t-x)!(n-t)!}p^x q^{t-x} (1-p-q)^{n-t}$ 
I think I could I figure out the algebra.. if I could just see what to do with the summation. I know I have to involve a t somewhere, but I'm not sure. Maybe somebody could just show me how to set up my summation correctly with the substitution? 
 A: Think about the probabilistic meaning of the trinomial distribution: there are $n$ independent trials, each trial has three possible results, say $a$, $b$ and $c$, whose respective probabilities are $p$, $q$ and $1-p-q$. Then $X$ counts the number of $a$ and $Y$ counts the number of $b$, among those $n$ trials. 
Hence $T=X+Y$ is the number of successes in $n$ i.i.d. trials, when success means $[a\ \text{or}\ b]$, hence has probability $p+q$. This is the binomial distribution $(n,p+q)$.
One can also finish your computation. You arrived at
$$
P[T=t]=\sum_{x=0}^{t} \frac{n!}{x!(t-x)!(n-t)!}p^x q^{t-x} (1-p-q)^{n-t},
$$
(note the sum up to $t$, not $n$), that is,
$$
P[T=t]={n\choose t}(1-p-q)^{n-t}\sum_{x=0}^{t} {t\choose x}p^x q^{t-x},
$$
or, with $r=p+q$,
$$
P[T=t]={n\choose t}(1-r)^{n-t}r^t.
$$ 
A: When the trinomial is expressed in the degenerate 3 variable $(X,Y,Z)$ form, we have the constraint $X+Y+Z=n$ which ties everything together. When using the trinomial in your 2 variable $(X,Y)$ form, it is important to include a constraint $X + Y \le n$. 
GIVEN: Random variables $(X,Y)$ have joint pmf $f(x,y)$: 

where the Boole term imposes the constraint $X + Y \le n$. 
Step 1: Consider the transformation $(T = X + Y, W = X)$. The joint pmf of $(T,W)$, say $g(t,w)$, can be obtained using the Method of Transformations, as:

where I am using the mathStatica function Transform to automate the calculation for me, though one could do it manually too, of course. (I should add that I am one of the authors of the package.) 
Step 2: To obtain the marginal pmf of $T=X+Y$, we just have to sum out $W$. Since $W = X$, and $T = X+Y$, and all our variables are positive, we clearly have $0 \le W \le T$, so the marginal pmf of $T$ is:
$$\sum _{w=0}^t g(t,w) =  \frac{n!}{t! (n-t)!} (p+q)^t (1-p-q)^{n-t}$$
with domain of support $t = {0,1, ..., n}$. The pmf of $T=X+Y$ is thus $Binomial(n,p+q)$. 
All done.
