There are 9 4-term factors in (w+x+y+z)9

(w+x+y+z)(w+x+y+z)(w+x+y+z)(w+x+y+z)(w+x+y+z)(w+x+y+z)(w+x+y+z)(w+x+y+z)(w+x+y+z)

To multiply it all the way out we would choose 1 term from each
factor of 4 terms.

To get w²x³yz³, 

we choose the 2 w terms from the 9 4-term factors.
That leaves 7 4-term factors from which to get the 3 x factors.
We choose the 3 x terms from the remaining 7 4-term factors,
That leaves 4 4-term factors from which to get the 1 y factor.
We choose all 3 z terms from the remaining 3 4-term factors,
and we're done.

So, going through it:

We can choose 2 of the 9 factors from which to get the two w's 
to make the w² in C(9,2) ways

For each of those C(9,2) ways, we can choose 3 of the remaining 7
factors from which to get the 3 x's to make the x³ in C(7,3) ways.

That's C(9,2)C(7,3) ways to get w²x³

For each of those C(9,2)C(7,3) ways, we can choose 1 of the remaining 4
factors from which to get the 1 y in C(4,1) ways.

That's C(9,2)C(7,3)C(4,1) ways to get w²x³y

For each of those C(9,2)C(7,3)C(4,1) ways, we can choose all 3 of the 
remaining 3 factors from which to get the 3 z's in C(3,3) ways, which
is of course just 1 way.

That's C(9,2)C(7,3)C(4,1)C(3,3) ways to get a product of w²x³yz³
choosing one letter from each of the 9 4-term factors.

So that's C(9,2)C(7,3)C(4,1)C(3,3) = 36*35*4*1 = 5040 ways.

So the coefficient of the term in w²x³yz³ is 5040.

Edwin