Counting Subgraphs of simple graphs Assuming $r\le n$
What is the number of subgraphs $K_r$ in $K_n$? I am not sure if this one is just $n \choose r$ or I have to divide that by 2?
What is the number of subgraphs $P_r$ in $K_n$? Is it just $\frac{n!}{2(n-r)!}$?
number of subgraphs $K_{r,r}$ in $K_{p,q}$? 
number of subgraphs $C_{2n}$ in $K_{n,n}$? 
number of subgraphs $C_{r}$ in $K_n$? I think this is $ \frac {n \choose r}{2} $  
 A: A subgraph $K_r$ of $K_n$ is completely determined by the set of $r$ vertices chosen, so there are indeed $\binom{n}r$ of them.
I’m assuming that $P_r$ is the path graph on $r$ vertices. Every sequence of $r$ of the vertices of $K_n$ determines a unique directed path, and there are $\frac{n!}{(n-r)!}$ such sequences, so there are $\frac{n!}{(n-r)!}$ directed paths. Each actual $P_r$ can be traversed in either of two directions, so you’re right: there are $\frac{n!}{2(n-r)!}$ subgraphs $P_r$ of $K_n$.
You can use similar reasoning to get the number of subgraphs $C_r$ in $K_n$. (I’m assuming that $C_r$ is an $r$-cycle.) If you delete any one edge from a subgraph $C_r$ of $K_n$, you get a subgraph $P_r$. Each subgraph $C_r$ has $r$ edges and therefore corresponds to $r$ subgraphs $P_r$. There are $\frac{n!}{2(n-r)!}$ subgraphs $P_r$ of $K_n$, so there are $\frac{n!}{2r(n-r)!}$ subgraphs $C_r$ of $K_n$. As a quick rough and ready sanity check, consider the case $n=4$ and $r=3$: it’s easy to verify that $K_4$ has $4$ subgraphs $K_3$, each obtained by omitting one vertex, and $\frac{4!}{2\cdot3(4-3)!}=4$. Note that your $\frac12\binom43=2$, so it definitely can’t be right.
Counting the subgraphs $K_{r,r}$ of $K_{p,q}$ is very similar to counting the subgraphs $K_r$ of $K_n$: it’s just a matter of counting the ways to choose the vertices to get a $K_{r,r}$ subgraph. See if you can do so, using the same kind of reasoning that tells us that $K_n$ has $\binom{n}r$ subgraphs $K_r$.
That leaves only the number of subgraphs $C_{2n}$ in $K_{n,n}$. List the vertices of $K_{n,n}$ as $u_1,u_2,\ldots,u_n$ in one part and $v_1,v_2,\ldots,v_n$ in the other part. A subgraph $C_{2n}$ must include every one of these vertices, so we might as well start at vertex $u_1$. How many choice are available for the next vertex in the cycle? What about the one after that? Keep going: the number of available choices changes in a very simple way. Once you know how many choices are available at each stage in the construction of a $2n$-cycle, you can easily combine those numbers to get the total number of $2n$-cycles in $K_{n,n}$ (though don’t forget that every cycle can be traversed in two directions).
