# how is the abs val of excess of p(n|odd # of parts) OVER p(n|even #of parts) = p(n|distinct odd parts?)

A question in The Elementary Theory Of Partitions asks the reader to show that the absolute value of excess of the number of partitions $$n$$ with an odd number of parts over the number of those with an even number of parts equals the number of partitions of $$n$$ into distinct odd parts.

I think that i have seen proof that $$|p_{even}(n) - p_{odd}(n)| = p_{do}(n)$$ but I do not believe that is what the question is asking, I think it wants: $$\left|\frac{p_{odd}(n)}{p_{even}(n)}\right|=p_{do}(n)$$ Is this what the question asks or am I misunderstanding? If it is true how is it true?

The “excess of $$A$$ over $$B$$” means $$A-B$$. In other words, the amount by which $$A$$ exceeds $$B$$.