Derangement and Recursive Relation Given $n$ different objects,they have to be arranged in such a way that $k$ objects won't occupy their initial position.
How to solve this problem? 
For example,if $n=7,k=3$ how many permutations will we get so that exactly $3$ objects won't occupy their initial position?
 A: In the case where $n=7$ and $k=3$, there are $\binom{7}{3}$ ways to choose the 3 objects which will be moved from their initial positions, and then there are only two ways to rearrange these 3 objects so that they are in different positions.  (If their initial order was ABC, they can be rearranged as BCA or CAB.)
Therefore the answer in this case is $\binom{7}{3}\cdot 2=70$.
In general, as indicated in Andre Nicolas's comment, the answer would be $\binom{n}{k}\cdot D_{k}$, 
where $D_k$ is the number of derangements of $k$ objects.
A: Do you familiar with inclusion-exclusion?
The principle is simple, define different attributes - in your case $n$ different, which the $i$-th imply that the $i$-th object is not in its initial place.
Then, you calculate how many permutations are there that satisfy $k$ specific attributes, and sum them up over all $k$ objects choices, now you substract those cases that satisfy $k+1$ since they were counted multiple time in the $k$ and continue in that way.
Further details http://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle 
