What does the expansion of (x + y + z)^n look like when n = 1, 2, 3? and also how does this relate to the question: Find the number of strings of 3 nonnegative integers whose sum is 13 which I cannot figure out.
Each term of $\displaystyle \left( {x + y + z} \right)^n$ look like $\displaystyle \frac{{n!}}{{a!b!c!}}x^a y^b z^c \,,\,\left( {a + b + c = n} \right)$.
As to your question “Find the number of strings of 3 nonnegative integers whose sum is 13 which I cannot figure out.” I hope that you have not confused two different ideas. The answer to that question is the value of the coefficient of $\displaystyle x^{13}$ in the expansion of $\displaystyle \left( {\sum\limits_{k = 0}^{13} {x^k } } \right)^3 $.
Hello, chadlyter!
The "Trinomial Expansion" . . . an interesting concept.
What does the expansion of $\displaystyle (x + y + z)^n$ look like when $\displaystyle n = 1, 2, 3$?
When I was in college, I expanded some trinomials and made some "discoveries".
It's best to write the terms in a triangle.
$\displaystyle (x + y + z)^1 \;=\;\begin{array}{ccc}& x & \\ \\ y & & z \end{array}$
$\displaystyle (x+y+z)^2 \;=\;\begin{array}{ccccc}& & x^2 & & \\ \\ & 2xy & & 2xz & \\ \\ y^2 & & 2yz & & z^2 \end{array}$
$\displaystyle (x+y+z)^3 \;=\;\begin{array}{ccccccc}& & & x^3 \\ \\ & & 3x^2y & & 3x^2z \\ \\ & 3xy^2 & & 6xyz & & 3xz^2 \\ \\ y^3 & & 3y^2z & & 3yz^2 & & z^3\end{array}$
$\displaystyle (x+y+z)^4 \;=\;\begin{array}{ccccccccc}& & & & x^4 \\ \\
& & & 4x^3y & & 4x^3z \\ \\
& & 6x^2y^2 & & 12x^2yz & & 6x^2z^2 \\ \\
& 4xy^3 & & 12xy^2z & & 12xyz^2 & & 4xz^3 \\ \\
y^4 & & 4y^3z & & 6y^2z^2 & & 4yz^3 & & z^4 \end{array}$
There are some remarkable patterns.
Reading down the left edge, we have: .$\displaystyle (x + y)^n$
Reading down the right edge, we have: .$\displaystyle (x+z)^n$
Reading across the bottom, we have: .$\displaystyle (y+z)^n$
What about the "inner" coefficients?
If we stack these triangles, we get "Pascal's Tetrahedron".
Every number is the sum of the three numbers above it.
However, we don't really need to carry around this tetrahedron.
For $\displaystyle (x+y+z)^n$, the coefficient of $\displaystyle x^ay^bz^c$ is, of course: .$\displaystyle {n\choose a,b,c} $