Find the number of terms in the expansion of $\displaystyle (x+y+z+2w)^{12}$
Find the term involving $\displaystyle x^2y^2z^4w^4$
The number of terms is equal the the number of non-negative integer solutions to the equation $\displaystyle x+y+z+w=12$.
So find $\displaystyle \binom{12+4-1}{12}$.
For the second part calculate $\displaystyle 2^4\frac{12!}{(2!)^2(4!)^2}$