Combinatorial proof of an identity involving integer partitions and their conjugates I have that $\lambda$ and $\mu$ are integer partitions and $\lambda^*$ and $\mu^*$ are their conjugates (respectively).  I am trying to use counting arguments to prove that:
\begin{align}
\sum_{i,j}min(\lambda_i,\mu_j) = \sum_k \lambda_k^*\mu_k^*
\end{align}
I am having trouble coming up with a model for what this counts.
 A: Let's use Young diagrams to visualize the partitions. For the ease of explanation, we will work with a specific example. However, it is not hard to see how to do the general case. The example we will work with is $\lambda = (3,2,1,1)$ and $\mu = (4, 1)$. Then the Young diagram for $\lambda$ looks like this: 

The young diagram for $\mu$ looks like this 

Now we take a three-dimensional coordinate system. We place the Young diagram for $\lambda$ under the $xy$-plane, and the Young diagram for $\mu$ behind the $xz$-plane. In the positive $xyz$-octant we now place cubes in all spots that are above a square of $\lambda$ and in front of a square of $\mu$. See the diagram below for a visualization. Here $\lambda$ is colored red, $\mu$ is colored blue, and the new cubes we added are colored yellow. 

Now if we would want to count the number of yellow cubes, there are several ways to proceed. One way would be to look at each horizontal line of cubes (i.e. in the left-to-right direction). If we number from the back and from below, the $i$-th line on the $j$-level contains exactly $\textrm{min}(\lambda_i, \mu_j)$ blocks. Hence, there are $\sum_{i,j} \textrm{min}(\lambda_i, \mu_j)$ cubes. 
On the other hand, we can look at the slices that are parallel to the $yz$ plane (i.e. the plane normal to both Young diagrams). Counting from left to right, the $k$-th slice is a rectangle with length $\lambda_k^*$ and heigth $\mu_k^*$. Hence, there are $\sum_{k} \lambda_k^*\mu_k^*$ yellow blocks. We conclude that $$\sum_{i,j} \textrm{min}(\lambda_i, \mu_j) = \sum_{k} \lambda_k^*\mu_k^*.$$
 
