How many copies of P3 are there in K10 How many copies of P3 are there in K10?
I can draw both of the graphs, but I don't know how you calculate this and assume there is a method that can be used to make this easier.
Thanks
I assume $P_3$ is the graph

       *---*---*


An embedding of this into $K_{10}$ is simply a labeling of it with three different numbers from $\{1,2,\ldots,10\}$.

To choose such a labeling, first pick the label of the middle node; this can be done in $10$ ways. Then pick the numbers that will label the two end nodes; this can now be done in $\binom92$ ways (since there's an automorphism of $P_3$ that swaps the ends).
Just realised both mine and hennings answers are wrong. Purely because $P_3$ has 4 nodes not 3. It has 3 edges not 3 vertices.

Labelling the nodes of $P_3$  as {$1,2,3$} then there are $3$ 'triangles minus one edge' that you can make. 

Using this you can say that for every $K_3$ in $K_{10}$ there are $3P_3$.

So the number of $K_3$ in $K_{10}$ is ($^{10}C_3$) which gives the number of $P_3$ as $3(^{10}C_3)=360=10(^9C_2)$ as Henning said.

$^{10}C_3$ is 10 choose 3