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How many different dozens of donuts can be made using at most 4 glazed, 3 chocolate, 5 jelly filled and 1 Bizmark?

Is we solve this problem by generating function

as my idea is Let $x_1$ be the glazed and $x_2$ be the chocolate , $x_3$ be the jelly $x_4$ be Bizmark

where $0 \leq x_1\leq 4$, $0 \leq x_2\leq 3$, $0 \leq x_3\leq 5$, $0 \leq x_4\leq 1$,

how can we processed further

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    You are seeking the coefficient of $x^{12}$ in $(1+x+x^2+x^3+x^4)(1+x+x^2+x^3)(1+x+...+x^5)(1+x)$. – 2017-02-17

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I guess you could solve using a generating function, but it seems to me that you can do this in a more straightforward manner. You have to choose to leave out one of one type of donut (as you only have 13 total to choose from), so there are 4 ways to do this.

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    I really like this answer because of how simple it is. Not to say that you can't solve it using a generating function, but this method could save a lot of time on calculations – 2017-02-17