Generating function - ways of distribution

Question

In how many different ways can eight identical cookies be distributed among three distinct children if each child receives at least two cookies and no more than four cookies?

$x_1 \ ^2$ + $x_2 \ ^2$ + $x_3 \ ^2$ = 8

could be written as -

($x^2$ + $x^3$ + $x^4$)($x^2$ + $x^3$ + $x^4$)($x^2$ + $x^3$ + $x^4$) = $(x^2 +x^3 + x^4)^3$

Now I had to find out coefficient of $x^8$. This was a small problem so I manually calculated it and I got the answer 6, but is there any direct way to find out $a_kx^k$? without any expansion. I know the same problem can be solved using stars and bars method, but I wanted to know if there is any direct way to find coefficient using above method or not.

Answer 1

A handy identity that often helps in this type of problem is $$1+x+x^2+ \dots +x^n = \frac{1-x^{n+1}}{1-x}$$ so let's see if we can apply it here: $$\begin{align} (x^2+x^3+x^4)^3 &= x^6 (1+x+x^2)^3 \\ &= x^6 \left( \frac{1-x^3}{1-x} \right)^3 \\ &= x^6 (1-x^3)^3 (1-x)^{-3} \\ &= x^6 (1-3x^3+3x^6-x^9) \sum_{i=0}^{\infty} \binom{3+i-1}{i} x^i \end{align}$$ We applied the Binomial Theorem for a negative exponent in the last step.

From the last expression we can easily see that the coefficient of $x^8$ is $$\binom{3+2-1}{2} = 6$$

Answer 2

In this case, it's very simple because the problem is small, as you say. We have $$(x^2+x^3+x^4)^3=x^6(1+x+x^2)^3,$$ so we need the coefficient of $x^2$in $(1+x+x^2)^3.$ By inspection, we must have either one factor of $x^2$ and two factors of $1$ ($3$ possibilities,) or one factor of $1$ and two factors of $x$ ($3$ possibilities,) for a total of $\boxed{6}.$

Of course, in larger problems, this won't work.