2) **0** is the category with no objects or morphisms (i.e., both sets are \\(\emptyset\\).

3) Claim: There are \\(\frac{n(n+1)}{2}\\) morphisms in **n** for all \\(n\in\mathbb{N}\\).

Proof: For every two objects, say \\(p>q\\), in **n** there is exactly one one morphism, that obtained from the compositions \\(f_{p-1}\circ f_{p-2}\circ\cdots\circ f_{q+1}\circ f_q\\). By definition of **n**, these constitute all of the non-identity morphisms in **n**. So, there are \\(\binom{n}{2}=\frac{n(n-1)}{2}\\) non-identity morphisms and \\(n\\) identity morphisms, for a total of \\(\frac{n(n+1)}{2}\\) morphisms.

3) Claim: There are \\(\frac{n(n+1)}{2}\\) morphisms in **n** for all \\(n\in\mathbb{N}\\).

Proof: For every two objects, say \\(p>q\\), in **n** there is exactly one one morphism, that obtained from the compositions \\(f_{p-1}\circ f_{p-2}\circ\cdots\circ f_{q+1}\circ f_q\\). By definition of **n**, these constitute all of the non-identity morphisms in **n**. So, there are \\(\binom{n}{2}=\frac{n(n-1)}{2}\\) non-identity morphisms and \\(n\\) identity morphisms, for a total of \\(\frac{n(n+1)}{2}\\) morphisms.