If $G$ is an abelian $p$-group and $p^k\le\exp(G)$, then the number of cyclic subgropups in $G$ is $$ {\rm c}_k(G)=\frac{|\Omega_k(G)-\Omega_{k-1}(G)|}{(p-1)p^{k-1}}. $$ If the type of $G$ is given, it is easy to compute $|\Omega_k(G)|$. The displayed formula is also suit for the regular $p$-groups. Computation of subgroups of given order is known for not very complicated $p$-groups, for example, for metacyclic $p$-groups (see \S 124 in the book of Berkovich-Janko; in that series a great number of counting theorems is proved, in particular, celebrated Kulakoff's theorem for all $p$).

If $G$ is an abelian $p$-group and $p^k\le\exp(G)$, then the number of cyclic subgropups of order $p^k$ in $G$ is $$ {\rm c}_k(G)=\frac{|\Omega_k(G)-\Omega_{k-1}(G)|}{(p-1)p^{k-1}}. $$ If the type of $G$ is given, it is easy to compute $|\Omega_k(G)|$. The displayed formula is also suit for the regular $p$-groups. Computation of subgroups of given order is known for not very complicated $p$-groups, for example, for metacyclic $p$-groups (see \S 124 in the book of Berkovich-Janko; in that series a great number of counting theorems is proved, in particular, celebrated Kulakoff's theorem for all $p$).