TITLE: How many ways can $10$ digits be written down so that no even digit is in its original position
QUESTION [5 upvotes]: If I have the numbers $0,1,2,3,4,5,6,7,8,9$ written down in that order, how many ways can the $10$ digits be written down so that no even digit is in its original position?
It would seem that I can move rewrite it starting from $0$ in $9!$ ways, and same with $2,4,6,8$, hence the answer is $5*9!=1814400$ is this correct?

REPLY [6 votes]: As suggested by Robert Israel, here is a solution using Inclusion-Exclusion:
Let S be the set of all arrangements of the digits, and for $1\le i\le5$
let $A_i$ be the set of arrangements with the $i$th even digit in its original position.
Then $|{A_1}^{c}\cap \cdots \cap {A_5}^{c}|=|S|-\sum|A_i|+\sum|A_i\cap  A_j|-\sum|A_i\cap A_j\cap A_k|+\cdots$
$=10!-\binom{5}{1}9!+\binom{5}{2}8!-\binom{5}{3}7!+\binom{5}{4}6!-\binom{5}{5}5!=2,170,680.$