In the normal derangement problem we have to count the number of derangement when each counter has just one correct house,what if some counters have shared houses. A derangement of n numbers is a permutation of those numbers in which none of the numbers appears in its original place. For example, the numbers {1,2,3} can be deranged into {2,3,1} and {3,1,2} But in our modified version no number in the derangement can be in the place that a number of the same value was in in the original ordering. So the numbers {1,1,2,2,3} could be deranged into {2,2,1,3,1}, {2,2,3,1,1}, {2,3,1,1,2}, and {3,2,1,1,2}.
How to go about it! any hints?
This is a permutation with forbidden positions. Your specific problem corresponds to the board below, where $X$ in square $(i,j)$ denotes that element $i$ is not allowed to go to position $j$.
XXOOO
XXOOO
OOXXO
OOXXO
OOOOX
Let $r_i$ denote the number of ways to place $i$ rooks on the $X$ squares, such that none attack each other.
$r_1=9$, because there are $9$ squares marked $X$.
$r_2=4^2+4+4+2+2$, because there are $4^2$ ways to pick one square from each of the $2\times 2$ grids, there are $4+4$ ways to pick one square from one of the $2\times 2$ grids plus the lower-right square, and there are $2+2$ ways to pick two squares from each of the $2\times 2$ grids.
You still need to find $r_3, r_4, r_5$. It's fairly tedious but possible. Then, the answer is $$5!-r_14!+r_23!-r_32!+r_41!-r_50!$$