Sum of coefficients of degree $r$ in the expansion of $(1 + x)^n (1 + y)^n (1 + z)^n$ In the expansion of $(1 + x)^n (1 + y)^n (1 + z)^n$, find the sum of the co-efficients of the terms of degree $r$.
As we don't have any info on $x,y,z$, putting $x=y=z$, we get the answer $^{3n}C_r$, but I am not satisfied.  
 A: Your approach is fine. To put it more formally we can introduce a variable $t$ and consider $[t^r]$, the coefficient of $t^r$ in
\begin{align*}
[t^r](1+tx)^n(1+ty)^n(1+tz)^n\tag{1}
\end{align*}

Since we want to find the sum of coefficients of degree $r$ in the expansion of $(1 + x)^n (1 + y)^n (1 + z)^n$ we set $x=y=z=1$ in (1) and obtain
  \begin{align*}
[t^r](1+tx)^n(1+ty)^n(1+tz)^n\big|_{x=y=z=1}&=[t^r](1+t)^{3n}\color{blue}{=\binom{3n}{r}}
\end{align*}

A: This solution can be linked to the Vandermonde's identity (https://en.wikipedia.org/wiki/Vandermonde%27s_identity).
You can think in a combinatory way:
You have $3$ groups with $n$ player each and you want to build a team with $r$ players, regardless the group they are from.
One way to do it is to group the $3$ groups in a big group with $3n$ players and pick your team, so you have $$\binom{3n}{r}$$ options.
The other way is to pick $a$ players from group 1, $b$ from group 2 and $c$ from group 3, so we get:
$$\sum_{a+b+c=r}\binom{n}{a}\binom{n}{b}\binom{n}{c}$$
And so we obtain the result.
Credits to my friend: Matheus Douglas
