<s>[INST] 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
 [/INST] 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.
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