Proposition 2.3. Let G be a connected graph of order at least two.(1) No graph of order two or more has th × c (G) = 1.(2) th × c (G) = 2 if and only if γ(G) = 1.((4) th × c (G) = 4 if and only if G satisfies one of the following conditions: (i) γ(G) = 2 and capt 1 (G) ≥ 3. (ii) c(G) = 1 and capt(G) = 3. Proof. (1) and (2) are immediate. (3): By factoring, we see that th × c (G; S) = 3 can be achieved only by |V (G)| = |S| = 3, or by |S| = 1 and capt(G; S) = 2. For a connected graph of order three, γ(G) = 1 and thus th × c (G) = 2. So th × c (G) = 3 implies c(G) = 1 and capt(G) = 2. For the converse, suppose c(G) = 1 and capt(G) = 2, so th × c (G) ≤ 3. Since c(G) = 1 and capt(G) = 2 implies γ(G) > 1, th × c (G) > 2. It is shown in the proof of Theorem 4.1 in [10] that the graphs for which c(G) = 1 and capt(G) = 2 are those described in (3i) and (3ii). (4): By factoring, we see that th × c (G; S) = 4 can be achieved only by |V (G)| = |S| = 4, by |S| = 2 and capt(G; |S|) = 2, or by |S| = 1 and capt(G; |S|) = 3. Any connected graph of order four has γ(G) ≤ 2. To ensure th × c (G) ≥ 4, we need capt 1 (G) ≥ 3. It is clear that either of the conditions γ(G) = 2 and capt 1 (G) ≥ 3, or c(G) = 1 and capt(G) = 3, implies th × c (G) = 4.It is immediate that th × c (K n ) = 2 and th × c (K 1,n−1 ) = 2 for n ≥ 2. A graph G is a chordal graph if it has no induced cycle of length greater than 3. The next result is less elementary than the previous ones.