I seem to have too many balls in the air at the moment. So let me drop one here. Any thoughts very welcome.
A k-hypergraph consists of a set X of vertices and a collection of k-element subsets called edges. It is k-partite if there is a partition of the vertices into k parts so that every edge is a transversal to this partition, which I will call a k-partition.
Clearly there is a tension between edges and k-partitions; you would expect that more k-partitions would imply fewer edges. In one special case this is true. The existence of a second k-partition implies the existence of two vertices x,y which lie in the same part of one of the k-partitions and different parts of the other; then no edge can contain x and y.
My conjecture is very specific:
Suppose that the k-hypergraph on n vertices has at least kr k-partitions, for some positive integer r. Then there is a k-partite k-hypergraph H1 on n−r vertices which has at least as many edges as H.
The hypotheses are not pulled out of the air. Adding an isolated vertex v to a k-partite k-hypergraph multiplies the number of k-partitions by k, since v can be added to any one of the parts. So the conjecture is true if the hypergraph has r isolated vertices.
The conjecture is true if k = 2. For a connected bipartite graph has a unique bipartition; so a graph with 2r bipartitions has at least r connected components, and a simple calculation gives the result.