What is the sum of the coefficients in the expansion of $(x+y+w+z)^{20}$

Question

Does the same method used to find sum of the coefficients for a binomial hold here?

Answer 1

That will be $(1+1+1+1)^{20} = 4^{20}$.

You just set $x=y=z=w=1$ in the formula, and simplify.