The notation $\lambda = (1^{m_1} 2^{m_2} 3^{m_3}) \vdash n$ means that $\lambda$ is a partition of $n$ with $m_1$ 1's, $m_2$ 2's and so on. For instance, the partition (3, 1, 1) of 5 can be written as $(1^2 3^1) \vdash 5$ and the partition (2, 2, 1) of 5 can be written as $(1^1 2^2) \vdash 5$.
The following identity appears on Page 10 of Igor Pak's survey on Partition Bijections as Problem (2.1.6). It is marked as being important to advance the understanding of the material.
Prove bijectively the following summation formula:
$$\sum_{\lambda = (1^{m_1} 2^{m_2} 3^{m_3}) \vdash n}{\frac{1}{1^{m_1} m_1 ! \; 2^{m_2} m_2 ! \; 3^{m_3} m_3 ! \cdots}} = 1$$
How can we show this?
The partitions correspond to conjugacy classes in the symmetric group $S_n$. To choose a permutation in the conjugacy class whose permutations have $m_k$ cycles of length $k$, we can choose elements for the cycles in $\frac{n!}{\prod_kk!^{m_k}}$ ways, we can choose their order within the cycles in $\prod_k(k-1)!^{m_k}$ ways, and we have to correct for the fact that permuting cycles of equal length among themselves in $\prod_km_k!$ ways yields the same permutation. Overall, that leads to a count of
$$ n!\prod_k\frac{(k-1)!^{m_k}}{k!^{m_k}m_k!}=n!\prod_k\frac1{k^{m_k}m_k!} $$
permutations in the conjugacy class. As there are $n!$ permutations in $S_n$ overall, summing over all partitions (and thus all conjugacy classes) and dividing by $n!$ yields your result.