interpreting partitions of integers(generating functions) the question is: In $ f(x) = \left (\frac{1}{1-x}  \right )\left (\frac{1}{1-x^2}  \right )\left (\frac{1}{1-x^3}  \right )$the coefficient of $x^6$ is 7. Interpret this result in terms of partitions of 6.
I checked the answer and it says the number of partitions of 6 into 1's, 2's and 3's is 7. But I'm having hard time understanding it. Can anyone explain this? An easy explanation would be very much appreciated..Thankyou

Writing each of the terms as power series, we have
$$\eqalign{f(x)
  &=(1+x+x^2+\cdots)(1+x^2+x^4+\cdots)(1+x^3+x^6+\cdots)\cr
  &=(1+x^1+x^{1+1}+\cdots)(1+x^2+x^{2+2}+\cdots)(1+x^3+x^{3+3}+\cdots)\ .\cr}$$
Now if you look carefully you will see that you can multiply terms, one from each bracket, to get an $x^6$ term in the following ways:
$$x^{3+3}\,,\ x^{1+2+3}\,,\ x^{1+1+1+3}\,,\ x^{2+2+2}\,,\ x^{1+1+2+2}\,,\ x^{1+1+1+1+2}\,,\ x^{1+1+1+1+1+1}\,.$$
These seven terms combine to give the coefficient of $7$ for $x^6$, and you can see that the exponents are all possible ways of writing $6$ as a sum of $1$s, $2$s and $3$s.
