Find the coefficient for a term in an expression We have the expression:

$$( 1 + x^1 + x^2 + x^3 + \dots + x^{27})(1 + x^1 + x^2 + \dots + x^{14})^2$$

For this expression how do you calculate the coefficient of $x^{28}$?
I know the answer is $224$, but I don't know how to calculate it. I know the methods could involve either using the multinomial coefficient, or factorizing with the geometric series reduction and then using synthetic division (which is longer). I wish to know the combinatoric approach please. 
 A: You are rolling three dice, two with $15$ sides labeled $0$ through $14$ and one with $28$ sides labeled $0$ through $27$.  You are asking how many ways there are to get a sum of $28$.  Note that no matter what the first two dice do, you can find exactly one throw of the third die to get a sum of $28$, unless the first two both come up zero. So there are $15^2-1=224$ combinations.
A: For $x\ne 1$, your expression is equal to
$$\frac{(1-x^{28})(1-x^{15})(1-x^{15})}{(1-x)^3}.$$
Expand $\dfrac{1}{(1-x)^3}$ using the generalized Binomial Theorem, or by differentiating the series $1+x+x^2+x^3+\cdots$ twice. Now the coefficient of $x^{28}$ is not difficult to compute. 
A: The idea is you want exponents to sum to $28.$
So $(1+x+\ldots + x^{27})(1+x + \ldots + x^{14})(1+x + \ldots + x^{14})$ . . . well, consider $a+b+c = 28$ with the restriction that $0\le a \le 27$ and $0\le b,c \le 14.$ Conveniently, enough this is equivalent to $a+ x = 28$ with $0 \le a \le 27$ and $1 \le x \le 28$ (since you need $a \le 27)$; obviously you'll have to count some solutions multiple times.
For each choice of $1 \le x \le 14,$ you will have precisely $x + 1$ possibilities for the ordered pair $(b,c),$ and for $15 \le x \le 28,$ you will have $29-x$ (this isn't difficult to count). Moreover, for each $x,$ there's exactly one $a$ that satisfies $a+x=28,$ so you just want the sum $\displaystyle\sum_{k=1}^{14} (k+1) + \sum_{k=15}^{28} (29-k) = 224.$
