Find coefficient of $w^2x^3y^2z^5$ when $(2w+x+3y+z)^{12}$ is expanded.

Question

For my combinatorics class, we're using combinations to compute the coefficient of a given term from an expression's expansion. In the book, there's an example with a trinomial, expanded from $(x+y+z)^9$ for the term $x^2y^3z^4$. Since $x$ is multiplied 2 times, the textbook uses $C(9,2)$, then since $y$ is multiplied 3 times, we get $9-2=7$ therefore, $C(7,3)$ and finally, by that logic $z$ would be $C(4,4)$ so 1. Then, the coefficient is given by $$C(9,2)*C(7,3)$$ I used the same method in the exercise mentioned in the title, with 4 variables. I got $$2C(12,2)*C(10,3)*3C(7,2) = 997920$$ The solution book's answer is $5,987,520$.

What am I missing?

Answer 1

Remember that the $2$ in front of $w$ and the $3$ in front of $y$ are both squared along with their variable. That should give you an additional factor $6$, and the answer $5\,987\,520$.

Answer 2

Hint: The multinomial theorem says that $$ (2w+x+3y+z)^{12}=\sum_{a+b+c+d=12}\binom{12}{a,b,c,d}(2w)^ax^b(3y)^cz^d. $$