How to find the coefficient of $x^{203}$ in the expansion of $(x-1)(x^2 - 2)(x^3-3)\dots(x^{20} - 20)$?

Question

How to find the coefficient of $x^{203}$ in the expansion of $(x-1)(x^2 - 2)(x^3-3)\dots(x^{20} - 20)$?

I took $x$ as common from each bracket making $x^{190}$ but I don't understand what to do next. Please help.

Answer 1

The ways of writing $7$ as a sum of distinct positive integers are $7,1+6,2+5,3+4,1+2+4$, so the coefficient of $x^{203}$ will be $-7+1\cdot 6 + 2\cdot 5 + 3 \cdot 4 - 1\cdot 2 \cdot 4 = 13$.

Answer 2

Hint. Note that $\sum_{k=1}^{20} k=210$ and $$(x-1)(x^2 - 2)(x^3-3)\cdots(x^{20} - 20)=x^{210} \left(1-\frac{1}{x}\right) \left(1-\frac{2}{x^2}\right) \left(1-\frac{3}{x^3}\right)\cdots\left(1-\frac{20}{x^{20}}\right).$$ Now consider the integer partitions with distinct parts of $210-203=7$: $$7,\quad 6+1,\quad 5+2,\quad 4+3,\quad 4+2+1.$$