Prove that the number of partitions of $2010$ into $10$ parts is equal to the number of partitions of $2055$ into $10$ distinct parts.

Question

Prove that the number of partitions of $2010$ into $10$ parts is equal to the number of partitions of $2055$ into $10$ distinct parts.

How we can prove this?My idea is to construct a bijection but I do not how we can do this.Please help me.

Answer 1

Take the Ferrers diagram for a partition of 2010 into 10 parts. Add 9 circles to the first row, 8 to the second row, ..., 1 to the next-to-last row. Now we have a partition of $2010+9+8+\cdots+1=2055$ into 10 parts, and they must all be distinct. This is the bijection you seek, now prove it's one-to-one and onto.

Answer 2

I am not sure completely but I think this is correct For every partition of 2010 into 10 parts order them in increasing order and to lowest number add $0$ to the next lowest add $1$ and so on. For example: Say the partition is $2+2+2+2+2+2+2+2+2+1992 $ it becomes $2+3+4+5+6+7+8+9+10+2001$. Due to this costruction every element in the set of partitions of 2010 has an image in the set of distinct partitions of 2055. Now we try to show that this is one one. So say $a_1+...+a_{10}$ and $b_1+b_2+..+b_{10}$ are both generated from the same partition of $2010$ the base partition is $a_1-0+...+a_{10}-9$=$b_1-0+b_2-1+..+b_{10}-9$ which implies that both the partitions are same. Onto is similarly done.