Writing a Sum of Partition Items in Combinatorial Form

Question

For each partition $\lambda$ we can define \begin{equation} n(\lambda) = \sum_{i \geq 1}(i-1)\lambda_i. \end{equation} According to my book this is equivalent to \begin{equation} n(\lambda)=\sum_{i \geq 1}{{\lambda_i'}\choose{2}} \end{equation} however I am struggling to prove this equality. Could anyone point me in the right direction?

Answer 1

HINT: Note that

$$\binom{\lambda_i'}2=\sum_{k=0}^{\lambda_i'-1}k=\sum_{k=1}^{\lambda_i'}(k-1)\;.\tag{1}$$

Now $\lambda_i'\ge k$ if and only if there are at least $k$ parts $\lambda_j$ of $\lambda$ such that $\lambda_j\ge i$. In terms of the Ferrers diagram of $\lambda$, the $i$-th column has $\ge k$ elements if and only if there are at least $k$ rows with $i$ or more elements. You want to show that

$$\sum_{i\ge 1}\binom{\lambda_i'}2=\sum_{i\ge 1}\sum_{k=1}^{\lambda_i'}(k-1)=\sum_{k\ge 1}(k-1)\lambda_k\;.$$