Prove that,
$$\binom{3n}{r}=\binom{3n-1}{r} + \binom{3n-1}{r-1}$$
Here is my solution below, this is my first time working through a pascal identity problem and I just wanted to see if this made sense:
By the definition $\binom{n}{r}$ we have,
$$\binom{3n-1}{r-1}=\frac{(3n-1)!}{(3n-1-(r-1))!(r-1)!} = \frac{(3n-1)!}{(3n-r)!(r-1)!}$$
and,
$$\binom{3n-1}{r}=\frac{(3n-1)!}{(3n-1-r)!r!}$$
Thus, starting with the right hand side of the equation,
$$\binom{3n-1}{r-1} + \binom{3n-1}{r} = \frac{(3n-1)!}{(3n-r)!(r-1)!} + \frac{(3n-1)!}{(3n-1-r)!r!} \\ = \frac{(3n-1)!r}{(3n-r)r!} + \frac{(3n-1)!(3n-r)}{(3n-r)!r!} \\ = \frac{(3n-1)!(r+3n-r)}{(3n-r)!r!} \\ = \frac{3n!}{(3n-r)!r!} \\ = \binom{3n}{r}$$ \
The proof looks perfect.
Why use $3n$ though? Note that the proof doesn't depend on using the $3$ at all. If you substituted $n$, you'd have the same proof for the general form of Pascal's identity.