Permutations and Derangements Determine the number of permutations of $\{1,2,...,9\}$ in which at least one odd integer is in its natural position. 
__ __ __ __ __ __ __ __ __ There are nine numbers to permute in the $9$ different position. There are $5$ odd numbers and $4$ even numbers. 
This is my idea- I'm not sure if it's right though. 
1) First I have to choose the odd number that will be in its natural position: $$\binom{5}{1}=5$$
2) Next I have to derange the other odd numbers: $$D_{4}$$
3) Lastly I have to permute the remaining numbers: $P(4,4)=24$
Total number of permutations:
$$=5D_{4}P(4,4)$$

You could also use rook polynomials to count the number of permutations where no odd integer is in its natural  position. Setting $N=9$ and $R(x)=(1+x)^5$, the  answer is 
$$\int_0^\infty x^N R(-1/x) \exp(-x)\,dx =205056.$$
Subtracting this from 9! gives 157824, the same as Ross's answer. 
