A partition of a positive integer $n$ is to write $n$ as a sum of some positive integers.. Show that the number of partitions of a positive integer $n$ into distinct parts is equal to the number of partitions of $n$ where all parts are odd integers.
Our goal is to establish a bijection between these two ways of partitions.
Given a partition of distinct parts, write every part in the form of $a\cdot 2^b$ where $a$ is an odd integer and $b$ is a non-negative integer. Then, we can split every part into $2^b$ of $a$. All parts will be odd integers. For example, given a partition of $15=10+4+1$, the following transformation can be performed to obtain an all-odd partition: $$\begin{align*} 15=\ & 10+4+1 \\=\ &5\cdot 2^1 + 1\cdot 2^2+ 1\cdot 2^0 \\=\ &(5+5)+(1+1+1+1)+(1)\\=\ &5+5+1+1+1+1+ 1 \end{align*}$$
On the other direction, we first group equal parts. Assuming there are $k$ parts of $a$ and $k=2^{k_1}+2^{k+2}+\cdots+2^{k_p}$ where $k_1 > k_2 > \cdots > k_p$. Then the sum of these parts can be written as $$k\cdot a = a\cdot 2^{k_1}+a\cdot 2^{k_2}+\cdots + a\cdot 2^{k_p}$$
Hence, it is possible to convert these parts to $a\cdot 2^{k_1}$, $a\cdot 2^{k_1}$, $\cdots$, $a\cdot 2^{k_p}$ which are all distinct. Different groups of such equal parts will result in different terms because $a\cdot 2^m$ will never equal $b\cdot 2^n$ if both $a$ and $b$ are different odd integers. For example, the partition $15=5+5+1+1+1+1+1$ can be transformed back to $10+4+1$ via $$\begin{align*} 15=\ &5+5+1+1+1+1+1\\=\ & 5\cdot 2 + 1\cdot 5 \\=\ &5\cdot(2^1) + 1\cdot(2^2 + 2^0)\\=\ & 5\cdot 2^1 + 1\cdot 2^2 + 1\cdot 2^0\\=\ &10+4+1 \end{align*}$$
Hence, there is indeed a bijection between these two cases. This means their counts are equal.