# is size of maximum matching less than that of maximum k-parity k-set packing?

My application problem can be modeled as maximum k-parity k-set packing problem, which is NP-hard. I know that maximum matching is polynomial time, hence I want to reduce k-parity k-set packing into bipartite maximum matching. I am wondering if there is any relation between their size.

Maximum k-parity k-set packing problem: We're given a universe $U$, which is a union of $k$ disjoint sets, $U = V_1 \cup \cdots \cup V_k$, and a family $\mathcal{S}$ of subsets of $U$, each $S$ contains exactly one vertex from each set, i.e., $S \in V_1 \times \cdots \times V_k$. A packing is a subfamily ${\mathcal {C}} \subseteq {\mathcal {S}}$ of sets such that all sets in ${\mathcal{C}}$ are pairwise disjoint. The size of that packing is $|{\mathcal {C}}|$. We are looking for the maximum-size packing.

In order to take advantage of polynomial time maximum matching we reduce $U$ to $U' = V_i \times V_j, i , j = 1, \cdots, k$ and $\mathcal{S}'$ be a family of subsets of $U$ such that each $S'$ contains exactly one vertex from $V_i$ and $V_j$. Let $d_{i,j}$ be the size of maximum matching and $d = \min_{i,j} \{d_{i,j} \}$.

For example in the figure below: $|C| = 2, d_{2,3} = d_{1,2} = d_{1,3} = 2$ hence $d = \min(2,2,2) = 2$.

My question: is this inequality true $|{\mathcal {C}}| \le d \le k \times |{\mathcal {C}}|$ ?

I think $|{\mathcal {C}}| \le d$ is obvious, do we also have $d \le k \times |{\mathcal {C}}|$?

Consider the case $k=3$, and the following set system: $$(a_1,b_1,c),\ldots,(a_n,b_n,c) \\ (a'_1,b',c'_1),\ldots,(a'_n,b',c'_n) \\ (a'',b''_1,c''_1),\ldots,(a'',b''_n,c''_n)$$ Every packing contains at most 3 sets, but $d=n+2$.