How many permutations of $$1,2,..., 9$$ are there that do not start or end with an even number?
This is my attempt
Condition 1 [Starts with even] => $$4 * 8!$$ Condition 2 [Ends with even] => $$4 * 8! $$ Condition 3 [Starts with even and Ends with even] => $$4*3*7!$$
Therefore, $$9! - (Condition1 +Condition2 -Condition 3)$$
Is this the right answer?
On
Five choices for first digit. Four remaining choices for last digit. Arbitrary order of the remaining 7 digits in between.
$5 \times 7! \times 4 = 100800$
Your (updated) method gives 100800. So we seem to agree.
Edited to reverse the transposition of "even" and "odd" in my solution.
Edited to incorporate edits to the problem's final formula.
On
Pick the first digit: ${5 \choose 1}$. Pick the last digit: ${4 \choose 1}$. Place the rest of them: $7!$.
That gives $5 \cdot 4 \cdot 7! = 100800$ possibilities.
On
As John pointed out, there's an easier approach to this, but you're nevertheless on the right track to a solution using inclusion-exclusion; you just missed a sign in your formula! If $B$ and $E$ are the sets of permutations beginning and ending with even numbers (respectively), then you want to compute $|B^c\cap E^c|$ where $S^c = S_9 - S$ denotes the complement of $S$ in the set $S_9$ of permutations on $1,\ldots,9$. So by inclusion exclusion and de Morgan's law
$$ |B^c\cap E^c| = |S_9 \setminus (B \cup E)| = 9! - (|B| + |E| - |B\cap E|) = 9! - |B| - |E| + |B\cap E| $$
On
Your solution is correct. It says
$$9! - (4 \cdot 8! + 4 \cdot 8! - 4 \cdot 3 \cdot 7!)$$
which calculates to
$$9! - 4 \cdot 7! \cdot (8 + 8 - 3) = 9! - 4 \cdot 7! \cdot 13 = (9 \cdot 8 - 4 \cdot 13) \cdot 7! = 20 \cdot 7!$$
Now other people's answers show another (simpler) way to see that the answer is $5 \cdot 4 \cdot 7!$.
Note: This answer pertains to the original version of the question, which did not have parentheses in the final displayed equation. end note
Your approach would be OK if you wrote it as
$$9!-Condition1-Condition2+Condition3$$
That would be using the Inclusion-Exclusion Principle: Start with the unrestricted number ($9!$), subtract off the numbers that violate the two conditions, but then realize you've subtracted twice the ones that violate both conditions.
But another way to solve the problem directly is to put odd numbers at the beginning and end, in $5\cdot4$ ways, and then fill in the rest in $7!$ ways.