1
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ Hint: The answer will be equal to 2 x (number of ways to arrange 500 (even) numbers) x (number of ways to arrange 500 (odd) numbers). And of course the 2nd and 3rd of those numbers are equal. $\endgroup$ – Paul Jan 21 '20 at 17:07
4
$\begingroup$

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 $2 \cdot (500!)^2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.