Math Question

Started by Da Pittman2 pages

Math Question

OK, I got a question and I hope that some math guru can help me out. I’m trying to figure out how many combinations can be made. You have an object that can have up to 10 options applied to it and can be any number of options so they could have none or all ten or any combination but no repeats. Any one knows how to figure this out?

ask ABS

shrug

way beyond me!

What's an option?

Google it?

Originally posted by FoxMeister
Google it?
I tried but but I don't know the name of the type of formula and everything I've tried so far comes up with nothing. If someone could give me the right key word that would help too.

Originally posted by Da Pittman
I tried but but I don't know the name of the type of formula and everything I've tried so far comes up with nothing. If someone could give me the right key word that would help too.

"Since all normal digits are used, there are 10000 possible combinations.
Calculate it by taking the number of possibilities in the first spot, 10,
times the number in the second (10), the third, fourth and so on: 10*10*10*10*10*10*10*10*10*10 =
10000000000"

awehuh Somethin like that

the answer is 7

Originally posted by FoxMeister
"Since all normal digits are used, there are 10000 possible combinations.
Calculate it by taking the number of possibilities in the first spot, 10,
times the number in the second (10), the third, fourth and so on: 10*10*10*10*10*10*10*10*10*10 =
10000000000"

awehuh Somethin like that

Thats right.

but if there cant be repeats then obviously it will be less because there will only be 9 possible numbers after the first one..

Originally posted by FoxMeister
"Since all normal digits are used, there are 10000 possible combinations.
Calculate it by taking the number of possibilities in the first spot, 10,
times the number in the second (10), the third, fourth and so on: 10*10*10*10*10*10*10*10*10*10 =
10000000000"

awehuh Somethin like that

Thanks, that is what I was thinking but I wasn't sure.

10
10squared
10cubed

ETC
ETC
ETC

Edit: Nevermind. Beat me to it. 😛

Math was my worst subject. hanuts I failed Algebra 1 twice.

I love math.

I made a thread recently in another forum where everyone challenges me with the craziest math problems just so I can test my math skills for future tests. I did pretty good. 😄

Originally posted by T.M
Thats right.

but if there cant be repeats then obviously it will be less because there will only be 9 possible numbers after the first one..

So that would be...

10*10*9*8*7*6...?

wouldn't it be 10!

Edit: which I think is 3.628.800

Originally posted by AbnormalButSane
wouldn't it be 10!

Edit: which I think is 3.628.800

Indeed...because it would be 10*9*8*7*6*5*4*3*2.

If no combinations can be repeated...it HAS to be this way.

Also: ABS, your brains make you even sexier.

Double edit- Pittman, are you writing a permutation program?

3.14 = pi.

Originally posted by dadudemon
Indeed...because it would be 10*9*8*7*6*5*4*3*2.

If no combinations can be repeated...it HAS to be this way.

Also: ABS, your brains make you even sexier.

Double edit- Pittman, are you writing a permutation program?

It is in a debate that I'm having about this game that I play and also I've always wondered my self how to caculate that but never asked until it came up now. I remember going over this in school but that was a LOOOOOOOOOOOOOOOOOONG time ago 😄
Originally posted by Impediment
3.14 = pi.
I love pie pitt_shifty

Originally posted by Da Pittman
It is in a debate that I'm having about this game that I play and also I've always wondered my self how to caculate that but never asked until it came up now. I remember going over this in school but that was a LOOOOOOOOOOOOOOOOOONG time ago 😄 I love pie pitt_shifty

Anytime is a good time for pi! 😱

math makes my head hurt