MHB Equinumerous partitions into distinct parts from X and Y?

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Here's this week's problem!

Problem: Let $X\subseteq \Bbb N$ with the property that $2j\in X$ for all $j\in X$. Let $Y = \{j\in X : j/2\notin X\}$. Show that the number of partitions of a positive integer $n$ into parts from $Y$ is equinumerous with the number of partitions of $n$ into distinct parts from $X$.Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

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No one answered this week's problem correctly. You can find my solution below.

Spoiler

Let $A(n)$ be the number of partitions of $n$ into parts taken from $Y$, and let $B(n)$ be the number of partitions of $n$ into distinct parts taken from $X$. Then

$$\sum_{n = 0}^\infty A(n)q^n = \prod_{n\in Y} \frac{1}{1 - q^n} = \prod_{n\in X - 2X} \frac{1}{1 - q^n},$$

and

$$\prod_{n\in X - 2X} \frac{1}{1 - q^n} = \prod_{n\in X} \frac{1}{1 - q^n} \prod_{n\in 2X} (1 - q^n) = \prod_{n \in X} \frac{1}{1 - q^n} \prod_{n\in X} (1 - q^{2n}) = \prod_{n\in X} \frac{1 - q^{2n}}{1 - q^n} = \prod_{n\in X} (1 + q^n) = \sum_{n = 0}^\infty B(n)q^n.$$

Therefore $A(n) = B(n)$ for all $n\in \Bbb N$.