So im trying to figure out a problem that goes like this.


I have 12 characters or spaces and each character can be either a 0 or a 1

eg. 000000000000, 000000000001, 010101010101

How many total combinations can there be without repeating?

I think the answer is 144, because i think the formula is X^y. Where X is the number of spaces, and y is the number of possible values for a specific character or space. Does that make sense? Any help before sunday would be greatly appreciated. Tanks

There will be 2^12 combinations, or 4096

EDIT: so yeah you were on the right track but you had the numbers switched around

Well it's more than that.

Your string is binary, convert it to decimal and 111111111111 is 4095 meaning there are 4096 possible

edit: Kyle90 is too fast

Wait. WITHOUT repetition? So you can't have all 1s or all 0s or any type of 1111... of binary string?

If so then your only combination is 1010101010101...

Right?

That has 13 digits, JG. (just being difficult :P)

If the repetition thing is handled that way, you could have 010101010101 or 101010101010.

But what's meant by no repeating? Girl-boy-girl-boy seating or that an entire sequence can't be repeated (rephrase as how many unique combinations are possible)? If the latter, then 4096 is still the standard.

But wait... according to masc wouldn't there be like...

2 ways?

Yeah, its how many unique combinations.


Its actually for a football bet on 12 teams where the payoff is like 2100:1. I was seeing if I would bet on every combination and make money. I guess not, maybe Ill try with 10.

Just so you know, if it was possible to make money from betting on everything everyone would do it.

Math?

That works for the lotto, though
A company a while back actually tried to get every single lotto combination. I remember that they didn't, I think that they ran out of funds, but they did hit the jackpot.

[QUOTE=Unknown User]Yeah, its how many unique combinations.[/QUOTE]

Then the rule is, for every binary number of N digits, there are 2^N possible unique values.

Isn't the general case (number of possible values in each digit)^(number of digits), so Hex would be 16^n, Oct 8^n, Dec 10^n, and so on?

Ja

For cases of repetition, yes.

For cases w/o repetition, it's n! [forall][isin=n]Z[/isin][/forall]

i have no idea what that means.

It means that JG likes to use fancy math symbols.

The upside-down "A" means "all", the "E"-looking thingy means "belonging to the set", and the "Z" is "the integers".