Let I = A ∪ B be a partially commutative alphabet such that two letters commute iff one of them belongs to A and the other one belongs to B. Let M = A ∗ × B ∗ denote the free partially commutative monoid generated by I. We consider the following six problems for rational (given by regular expressions) subsets, X, Y of M: (Q1): X∪ Y=0? (Q2): X⊆ Y? (Q3): X= Y? (Q4): X= M? (Q5): M− X finite? (Q6): X is recognized? It is known (see (Berstel, 1979)) that all these problems are undecidable if Card A > 1 and Card B > 1, and they are decidable if Card A = Card B = 1 (Card U denotes the cardinality of U). It was conjectured (see (Choffrut, 1986, p. 79)) that these problems are decidable in the remaining cases, where Card A = 1 and Card B > 1. In this paper we show that if Card A = 1 and Card B > 1, then the problem (Q1) is decidable, and problems (Q2)–(Q6) are undecidable. Our paper is an application of results concerning reversal-bounded, nondeterministic, multicounter machines and nondeterministic, general sequential machines.