An integer partition of (n) is a collection of parts summing up to \(n\). Moreover
Definition
Let \(p(n \mid \text{condition})\) denote the number of integer partitions of a given number that satisfy some condition.
One of the basic identities is
Partitioning Identity (2.2)
$$
\begin{align}
p(n \mid \text{even parts}) = p(n \mid \text{even number of each part})
\end{align}
$$
That is, the number of integer partitions made of even parts is the same as the number of integer partitions such that we have an even number of each part.
Example 1
Take \(n = 8\). Then using only even parts, we can write
$$
\begin{align}
8 &= 8 \\
8 &= 6 + 2 \\
8 &= 4 + 4 \\
8 &= 4 + 2 + 2 \\
8 &= 2 + 2 + 2 + 2.
\end{align}
$$
While partitions that have an even number of each part are
$$
\begin{align}
8 &= 4 + 4 \\
8 &= 2 + 2 + 2 + 2 \\
8 &= 3 + 3 + 1 + 1 \\
8 &= 2 + 2 + 1 + 1 + 1 + 1 \\
8 &= 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1
\end{align}
$$
A Natural Bijection
One way to show the identity above works is to come up with a bijection between the two sets. Suppose we’re starting from the set of integer partitions such that the parts are even. Then a natural thing to do is to split each part into two and keep splitting until no even parts remain.
$$
\begin{align}
8 &\rightarrow 4 + 4 \\
4 + 4 &\rightarrow 2 + 2 + 2 + 2\\
2 + 2 + 2 + 2 &\rightarrow 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1
\end{align}
$$
The end result will have an even number of parts. The inverse of this map is to take every pair of equal parts and merge them until no pairs remain equal so
$$
\begin{align}
1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 &\rightarrow 2 + 2 + 2 +2 \\
2 + 2 + 2 + 2 \rightarrow 4 + 4 \\
4 + 4 \rightarrow 8
\end{align}
$$
Let’s try this method on another case, take
$$
\begin{align}
6+2 &\rightarrow 3 + 3 + 1 + 1
\end{align}
$$
and the inverse map is
$$
\begin{align}
3 + 3 + 1 + 1 &\rightarrow 6 + 2
\end{align}
$$