<s>[INST] Show the coefficient of $x^3$ in the expansion of $(1+x+x^2+x^3)(1+x^2)(1+x^3)$ is the same as the number of partitions of $3$ (let a partition of $n \in \Bbb N, n>0 $ be a sum of positive integers)
How would you show that the number of different partitions of $3$ is the coefficient of 
$x^3$
in the
expansion of:
$(1+x+x^2+x^3)(1+x^2)(1+x^3)?$
My attempt:
since there are $3$ brackets to choose factors of $x^3$ from, we can say the following:
$ x^3 = x^a x^b x^c, a \in \{0,1,2,3\}, b \in \{0,2\}, c \in \{0,3\}$
The coefficient of $x^3$ in the expansion will equal the number of solutions satisfying $ x^3 = x^a x^b x^c$.
$a=0 \Rightarrow b=0 $ and $ c=3$,
$a=1 \Rightarrow b=2 $ and $ c=0$,
$a=2 \Rightarrow$ no solution.
$a=3 \Rightarrow b=0 $ and $ c=0$,
$\Rightarrow$ the coefficient of $x^3$ is $3$ which equals the number of partitions of $3$.
Do I understand what the question is asking? Because I'm wondering if just multiplying the brackets and obtaining the coefficient is an equivalent solution to the question.
Thanks.
 [/INST] You can see the "situation" as I try and depict in this sketch  
 
For what concerns the partitions of $3$ (and of $2$ and $1$) , all the necessary "pieces" are provided. 
Explanation of the sketch
When you expand the first polynomial, this will provide a factor $x^k$ to multiply the rest, and $k$ is either $0$ or $1$ or $2$ or $3$.
The second polynomial will provide a factor $x^j$ with $j=0,2$, and the third a factor $x^n$ with $n=0,3$.
When then you collect the terms $x^{(j+k+n)}$, grouping together all the terms with the exponent $j+k+n=q$, then each of the various terms will correspond to each one of the partitions of $q$ composed of $0..3$ parts $=1$ +$0..1$ parts $=2$ + $0..1$ parts $=3$, and the resulting coefficient of $x^q$ will be the number of those partitions.
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