How many ways are there to permute the integers from 1 to 1000 under the condition that two consecutive numbers must have different parity?

Question

How many ways are there to permute the integers from 1 to 1000 under the condition that two consecutive numbers must have different parity?

I know that there are two possible cases: either the sequence starts with an even or it is starts with an odd. Then the sequence alternates parity all the way through.

But I am not sure how to mathematically count the permutations of evens and permutations of odds. Like what formula do I use?

Answer 1

Of the $1000$ integers from $1$ to $1000$, half (i.e. $500$) of them are even and the other half are odd. Also, one only needs to know three things: the permutation of the even numbers, the permutation of the odd numbers, and whether the first term is even or odd. The answer is therefore $2 \cdot (500!)^2$.