Sum of coefficients of degree $r$ in the expansion of $(1 + x)^n (1 + y)^n (1 + z)^n$

Question

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.

Answer 1

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

Answer 2

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*}