Question about permutations

Joe · science forum beginner Joined: 25 Jun 2005 Posts: 22

Can anyone steer me in the right direction ? My question is, how many
permutations can I expect when I only use the odd or even numbers? For
example, 5!=120, but if I only use the permutations that are odd (ie,
1,3,5 ), I get 6. As I move up the number line, I expect to get more and
more, but not as many as N!. Is there a formula that would tell me this?

Thanks,
joe

Nigel · science forum beginner Joined: 03 Jun 2005 Posts: 37

Joe wrote:

Dan Akers · science forum addict Joined: 19 Jul 2005 Posts: 56

Joe wrote:
"Can anyone steer me in the right direction ? My question is, how many
permutations can I expect when I only use the odd or even numbers? For
example, 5!=120, but if I only use the permutations that are odd (ie,
1,3,5 ), I get 6. As I move up the number line, I expect to get more and
more, but not as many as N!. Is there a formula that would tell me
this?"
_____________________________________
Re;
Hi Joe. I'm hoping I've understood your question... Start with the
fact that there are 10 possible digits. If you use only the odd
"digits"; that cuts the number to 5. Thus, for a number consisting of N
digits, there are 5^N permutations. Thus for a 5 digit number, using
only odd digits, there are 3125 permutations.
On the other hand, if you meant that the number itself is to be "odd",
then the number of permutations changes to 10^(N-1)*5. Thus, for a 5
digit number, there would be 50,000 permutations. I hope that helps...

-Dan Akers

Joe · science forum beginner Joined: 25 Jun 2005 Posts: 22

"Dan Akers" <digikey@webtv.net> wrote in message
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