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:
$$=5D_{4}P(4,4)$$

The number of objects of $S$ which have at least one of the properties $P_1,P_2,...,P_m$ is given by $$|A_1\cup A_2\cup \cdots \cup A_m|=\sum|A_i|-\sum|A_i \cap A_j|+\sum|A_i\cap A_j\cap A_k|-\cdots +(-1)^{m+1}|A_1\cap A_2 \cap \cdots \cap A_m|.$$
If we select one of the odd integers to be in its natural position we can do so in ${5\choose 1}=5$ ways and then permute the remaining $8$ numbers in $8!$ ways for a total of $5\times 8!$ ways. If we select two odd integers to be in their natural positions we can do so in ${5\choose 2}=10$ ways and then permute the remaining $7$ integers in $7!$ ways for a total of $10\times 7!$ ways. If we select three odd integers to be in their natural positions we can do so in ${5\choose 3}=10$ ways and then permute the remaining $6$ integers in $6!$ ways for a total of $10\times 6!$ ways. If we select four odd integers to be in their natural position we can do so in ${5\choose 4}=5$ ways and permute the remaining $5$ integers in $5!$ ways for a total of $5\times 5!$ ways. If we select five odd integers to be in their natural position we can do so in ${5\choose 5}=1$ ways and then permute the remaining $4$ integers in $4!$ ways for a total of $4!$ ways. Thus there are $$5\times 8!-10\times 7!+10\times 6!-5\times 5! +4!=157824$$ ways for their to be at least one odd integer in its natural position.
