Krdytkyu

Let $E$ be the set of the first $x$ even numbers and $O$ the set of the first $y$ odd numbers.

(1) How many permutations are there of the set $E cup O$? I think that's just $(x+y)!$.

(2) How many permutations are there such that the last element of $E$ appears at an odd position?

Examples.

$[1,2,3,4,5]$ doesn't count since $4$ occurs at position $4$.

$[4,1,2]$ counts since $2$ is at position $3$.

$[2,1,5,3]$ counts since $2$ is at position $1$.

Original post:

Suppose we have an array of length z. Which consists of x different even numbers and y different odd numbers. So $x in [2_1,4_2,6_3 dots (2n)_x]$ and $y in [1_1,3_2,5_3 dots (2n+1)_y]$

So z = x+y.

We also know that x => 1 always applies.

1st question: How many different arrays of length z there are. That should be exactly $z!$ many or?

2nd question: How many different arrays are there so that the last even number in the array is on an odd index. (The array in this example starts with index 1)

Examples:

1) [1,2,3,4,5] This array has length 5. The last even number in the array is 4 and has index 4 in the array, so we don't count such an array.

2) [52,3,14]. The last even number in this array is 14 and has index 3. So such an array counts towards it.

3) [52,3,5,7]. The last even number in this array is 52 and has index 1. So such an array counts towards it.

Ok after reading the comment I believe you want to count (in your first question) arrays of length $z$ whose entries are the first $x$ even numbers and the first $y$ odd numbers, in some order, for some $x, y$ that add up to $z$.

Now for each value of $x$ the set of available numbers is fixed, and the number of orders in which these numbers can be written is indeed $z!$ as you write. However, picking a different value of $x$ we get a different set of numbers and hence another $z!$ arrays. If we exclude the cases where $x = 0$ or $y = 0$ there are $(z-1)z!$ different arrays: each of the $z - 1$ possibilities for $x$ (1, 2, ldots, z-1) yields $z!$ arrays. I'll illustrate this for $z = 5$.

$x = 1$ gives the set ${2, 3, 5, 7, 9}$ which gives rise to 120 arrays

$x = 2$ gives the set ${2, 4, 3, 5, 7}$ which gives rise to another 120 arrays

$x = 3$ gives the set ${2, 4, 6, 3, 5}$ which gives rise to another 120 arrays

$x = 4$ gives the set ${2, 4, 6, 8, 3}$ which gives rise to another 120 arrays

So for $z = 5$ we have $(z-1)z! = 480$ different arrays. If we do allow arrays in which all numbers are even or all numbers are odd $x = 0$ and $x = z$ are added to the team and the total number of arrays becomes $(z+1)z! = (z+1)!$

If you can tell me whether I understood the question correctly, I will edit in the answer to the second question.

EDIT: the edit of your question appeared while I typed this. Obviously the editor interpreted your question 1 in a different way from what I did. For that version of the question your answer $(x + y)!$ is of course correct.

EDIT 2: I'll type the answer to question 2 in the interpretation of the edited post. To get to the answer in my original interpretation you can simply multiply with $z-1$, as was the case with question 1 as well.

So... Let $(a)_b$ denote the product of $b$ subsequent descending numbers starting at $a$, so $(10)_3 = 10 cdot 9 cdot 8$, $n! = (n)_n$ etc. This will simplify notation down the road.

The last even number must occur in position $x$ or higher. If $x$ and $y$ are both even there are $y/2$ odd numbered position that satisfy this constraint: positions $x+1, x + 3, ldots, x + (y-1)$. If $x$ is even and $y$ is odd there are $(y+1)/2$ odd numbered positions available for the last even number: $x + 1, x+3, ldots, x+y$. If $x$ is odd and $y$ is even there are $y/2 + 1$ available spots: $x, x+2, ldots, x+y$ and when $x$ is odd and $y$ is odd there are $(y+1)/2$ available spots. Anyway: let's call the set of possible positions of the last even number $K$.

For each number $k in K$ we can count the number of possible permutations where the last even number lands in $k$ as follows.

EDIT 3: the following sentences have changed.

There are $x$ even numbers that can be the 'last' one, i.e. end up in position $k$. Then there are $x-1$ even numbers that are somehow distributed over positions $1, ldots, k-1$. This means that number 2 can choose from $k-1$ positions, after which number $4$ has $k-2$ positions left to choose from etc, yielding $(k-1)_{x-1}$ possibilities for the non-last even numbers. The $y$ odd numbers can then be put in the $y$ remaining spots in each of the conceivable $y!$ orders. The end result is hence:

$$sum_{kin K} x (k-1)_{x-1} y!$$

different arrays. I don't think this sum can be simplified much further, but someone might prove me wrong.