Da Pittman
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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?
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May 7th, 2008 07:23 PM
~Wålshy~
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ask ABS
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May 7th, 2008 07:24 PM
gefallen_engel
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way beyond me!
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May 7th, 2008 07:25 PM
It's xyz!
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What's an option?
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Last edited by Raz on Jan 1st 2000 at 00:00AM
May 7th, 2008 07:25 PM
FoxMeister
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Google it?
May 7th, 2008 07:26 PM
Da Pittman
"Pitt Happens"
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quote: (post ) 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.
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May 7th, 2008 07:27 PM
FoxMeister
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quote: (post ) 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"
Somethin like that
May 7th, 2008 07:30 PM
~Wålshy~
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the answer is 7
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May 7th, 2008 07:34 PM
T.M
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quote: (post ) 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"
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..
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May 7th, 2008 07:35 PM
Da Pittman
"Pitt Happens"
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quote: (post ) 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"
Somethin like that Thanks, that is what I was thinking but I wasn't sure.
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May 7th, 2008 07:36 PM
JacopeX
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10
10squared
10cubed
ETC
ETC
ETC
Edit: Nevermind. Beat me to it.
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(MEMBER WAS BANNED FOR THIS POST) The more you question it, the less you will understand.
May 7th, 2008 08:56 PM
Kelly_Bean
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Math was my worst subject. I failed Algebra 1 twice.
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May 7th, 2008 09:01 PM
JacopeX
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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.
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(MEMBER WAS BANNED FOR THIS POST) The more you question it, the less you will understand.
May 7th, 2008 09:03 PM
Da Pittman
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quote: (post ) 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...?
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May 7th, 2008 09:04 PM
AbnormalButSane
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wouldn't it be 10!
Edit: which I think is 3.628.800
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May 7th, 2008 11:27 PM
dadudemon
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quote: (post ) 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?
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May 8th, 2008 12:51 AM
Impediment
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3.14 = pi.
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May 8th, 2008 12:58 AM
Da Pittman
"Pitt Happens"
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quote: (post ) 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 quote: (post ) Originally posted by Impediment
3.14 = pi.
I love pie
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May 8th, 2008 01:00 AM
gefallen_engel
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math makes my head hurt
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May 8th, 2008 03:16 AM
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