# Proof of partitions in combinatorics

Problem: Prove that p(n) is equal to the number of partitions of the integer 2n with no odd parts.

Now i don't really understand what no odd parts means? I've looked online and have yet to find a problem saying no odd parts. Does this just mean even? I'm not sure where to start with this one due to the wording. Any help?

Also we arent allowed to use generating functions to show this

• 8=2+6 is a partition with no odd parts. 8=3+3+2 has odd parts namely 3. – Maesumi Mar 16 '14 at 19:17

## 2 Answers

The partition $10=6+2+2$ has no odd parts. All the entries are even.

The partition $10=4+3+3$ has some odd parts.

Note that there is a natural one to one correspondence between partitions of $n$ and partitions of $2n$ with no odd parts. Take any partiton of $n$ and double all the entries. We get a partition of $2n$ with no odd parts. and all such partitions can be obtained in this way.

• awesome thank you! is there an exact way to prove it, or can i say if one were to double all entries in partitions of n they would end up with partitions of 2n containing all even parts – Metzky Mar 16 '14 at 19:30
• You can be more formal. Let $\Pi_n$ be the set of all partitions of $n$, and let $\Pi_{2n}$ be the set of all partitions of $2n$ into even parts. Define the mapping $\varphi$ from $\Pi_n$ to $\Pi_{2n}$ by saying that for any partition $\pi$ of $n$, $\varphi(\pi)$ is obtained by doubling all the entries of $\pi$. This is obviously a one-to-one mapping. The mapping $\varphi$ is onto, for given any partition $\sigma$ of $2n$ into even parts, we have $\varphi(\pi)=\sigma$, where $\pi$ is obtained by halving all the entries of $\sigma$. But maybe this is overkill. – André Nicolas Mar 16 '14 at 19:44

No odd parts means that we cannot use odd numbers, only even ones. So a partition of $8$ that is allowed is $2+4+2$, but not $1+3+4$.

Now if we have a partition of $2n$ with only even parts, we could divide all numbers in them by 2, and if we have any partition of $n$ we can double the numbers...

• thats what i figured, just found it odd that they used no odds instead of even. thanks for the help! and thats what i thought for the proof, just trying to figure out how to put it on paper – Metzky Mar 16 '14 at 19:22