Let a partition be called self-transpose if its Ferrers diagram is symmetric about the diagonal, i.e. if the partition is its own transpose. For instance, there are two self-transpose partitions of 8: 4 + 2 + 1 + 1 and 3 + 3 + 2. There are also two ways to partition 8 into distinct odd parts: 7 + 1 and 5 + 3. These two quantities are explored below.
The problem I am looking at wants me to find 3 self transposes of the partition of 13 and the 3 partitions of 13 into distinct odd parts.
EDIT: So from @Greg Martin's answer I have gotten that the self-transposes are 4+4+3+2, 5+3+3+1+1, and 7+1+1+1+1 and the odd partitions are 13, 9+3+1, and 7+5+1. Now the question asks the following
Potentially using the above examples as a case study, give a combinatorial argument that there for any positive integer n, there are the same number of partitions of n into selftranspose parts and into distinct odd parts (hint: consider features in the Ferrers diagrams of the self-transpose partitions).
The reason for this is that if you look at the Ferrers diagram for the transposes, count each layer, and add them, you get the odd parts of the same partition. So every transpose can be written as an odd part of the same partition.
