practice questions not Homework
I have problems with this questions that I have answers for but cant understand how the answer was derived.
Q.1. In how many ways can the integer $1,2,3,...10$ be arranged in a line so that no even integer is in its natural position?
Answer) $10!-(5C1)9!+(5C2)8!......-(5C5)5!$ I am looking for some intuition as to why this and why isnt it just a $d_5$ (derangement of 5 numbers as we are dealing with 5 even numbers)
The number of derangements of five elements would tell you what order you were putting the even numbers in... but, not even derangements are enough to satisfy these conditions.
Consider, for instance, the ordering 10, 6, 8, 2, 4. This is a derangement of 2, 4, 6, 8, 10; however, where are we going to put the odd numbers? If you choose to put them in as 10, 1, 3, 5, 7, 6, 9, 8, 2, 4 then the result has an even integer (6) in natural position despite choosing the derangement!
So, instead, they start by counting the $10!$ ways to arrange the numbers with no restrictions, then work to subtract off the number of such arrangements which violate the condition. Etc.
I don't think that your formula, as written, can possibly be correct; why would you be choosing 9 things out of 5? And why would the next "logical" term by $\binom{5}{2}$? Double-check that.