you see the system $ x (mod m) = r,x(mod n) = s$has a solution if $ (m,n) = d$devides $ r-s$. if you make a bipartite graph with respect to two reduced residue system for $ m$ and $ n$,and join $ r$ from the first part to $ s$ from second part if $ r-s$ is devisible by $ d$.by inclusion principle we conclude that every part is regular,so we have a matching,using hall theorem.

can we make some bounds for the number of matchings?

in fact we can compute the exact number of matchings.

Also a generalization for this for $ m,n,k$ which $ \varphi(m)=\varphi(n)=\varphi(k)=c$ there exist $ m,n,k$ such that $ a_{1},a_{2},\dots,a_{c}$ such that is reduced residue system for $ m,n,k$. As I know the problem is not correct for more than three numbers.

dashmiz wrote: you see the system $ x (mod m) = r,x(mod n) = s$has a solution if $ (m,n) = d$devides $ r - s$. if you make a bipartite graph with respect to two reduced residue system for $ m$ and $ n$,and join $ r$ from the first part to $ s$ from second part if $ r - s$ is devisible by $ d$.by inclusion principle we conclude that every part is regular,so we have a matching,using hall theorem. I think this graph will be union of $ \phi (d)$ complete bipartite graphs, in each component there will be $ \frac{\phi (m)}{\phi (d)}$ =$ \frac{\phi (n)}{\phi (d)}$=$ v$ vertices in each vertice class. Even without Kőnig-Hall thm we can find $ v!^{\phi(d)}$ matchings. Correct me please if I was wrong...

Can we solve the problem without using graph ?