How many non-negative integer solutions are there for $x_1+x_2+...+x_5 = 20$ when $x_1 = 2x_2$

Question

$ x_1 + x_2 + x_3 + x_4 + x_5 = 20 $

Where

$ x_1 = 2x_2 $

So basically, I just substituted $ 2x_2$ into $x_1$ and the new equation becomes:

$ 3x_2 + x_3 + x_4 + x_5 = 20 $

$ x_3 + x_4 + x_5 = 20 - 3x_2 $

Now I will replace $x_2$ with simply $i$

$ x_3 + x_4 + x_5 = 20 - 3i $

Then the equation can be solved for all non-negative $i$ using the Distribution theorem: $C(r+n-1, r)$ with the general form:

$C( (20-3i)+3-1, 20-3i )$

So for $i={0, 1, 2, 3, 4, 5, 6}$:

$ \binom{22}{20} + \binom{19}{17} +\binom{16}{14} + ... + \binom{4}{2}$

Would this be correct?