How many permutations of [8] have neither 1 nor 2 as fixed points? I am attempting to understand the probleme des recontres and the principle of inclusion and exclusion. My solution for the question would be:
Use ${n \choose k}$ $D_{n-k}$ where D represents the number of derangements (fixed points). Then since there are 8 fixed points we have ${10 \choose 8}$ $D_{10-8}$. 
Am I approaching this properly? 
 A: There is no reason to look at derangements. The simplest approach is a short inclusion-exclusion calculation to count the permutations that do have $1$ or $2$ (or both) as a fixed point; this number is then subtracted from $8!$, the total number of permutations of $[8]$.
There are $7!$ permutations of $[8]$ having $1$ as a fixed point, and there are $7!$ permutations of $[8]$ having $2$ as a fixed point. However, there are $6!$ permutations of $[8]$ having both $1$ and $2$ as fixed points, so there are $2\cdot7!-6!=13\cdot6!$ permutations having $1$ or $2$ as a fixed point. Since there are $8!=56\cdot6!$ permutations altogether, there are $(56-13)6!=43\cdot6!=30960$ permutations for which neither $1$ nor $2$ is a fixed point.
It’s also possible to calculate the desired number directly.
If you interchange $1$ and $2$, you can do anything to the other $6$ numbers; that gives you $6!$ permutations. If you send $1$ to $2$ and $2$ to some $k\ne 1$, there are $6$ choices for $k$, and the numbers $3$ through $8$ can be sent to any permutation of $[8]\setminus\{2,k\}$; that’s another $6\cdot6!$ permutations. Similar reasoning shows that there are $6\cdot6!$ permutations that send $2$ to $1$ and $1$ to some $k\ne 2$. Finally, there are $6\cdot5$ ways to send $1$ and $2$ to elements of $[8]\setminus[2]$, and $6!$ ways to dispose of the remaining $6$ elements. The grand total is therefore $43\cdot6!=30960$ permutations for which neither $1$ nor $2$ is a fixed point.
