# Number Problem

• May 30th 2006, 08:46 PM
Maths Illiterate
Number Problem
Hi I Was Wondering How I Would Go About Geting All Possible Combinations For Numbers 1 Through To 8, I Know I Sound Real Dumb But Maths Isn't My Forte, At My School If Someone Gets It They Win \$200 ^^

Any Help Would Be Appreciated
• May 31st 2006, 09:33 AM
malaygoel
Quote:

Originally Posted by Maths Illiterate
Hi I Was Wondering How I Would Go About Geting All Possible Combinations For Numbers 1 Through To 8, I Know I Sound Real Dumb But Maths Isn't My Forte, At My School If Someone Gets It They Win \$200 ^^

Any Help Would Be Appreciated

Can you give an example of what kind of combination you are talking about?
• May 31st 2006, 05:37 PM
Maths Illiterate
11111111
11111112
11111113

......
88888886
88888887
88888888

That Kind Of Thing
• May 31st 2006, 06:25 PM
ThePerfectHacker
Quote:

Originally Posted by Maths Illiterate
11111111
11111112
11111113

......
88888886
88888887
88888888

That Kind Of Thing

Simple.

For each slot you have eight possibilities.
And there are eight slots thus you have.
\$\displaystyle 8^8=16777216\$.

I really do not see what is so challenging about this problem that the schools would offer \$200 dollars for anyone who can answer it (in fact the school can reject this deal because in law a contract is valid when there is consideration-meaning both parties gain something, which is not true here, thus no contract-basically no \$200). Unless it is a special-ed school :D
• Jun 23rd 2006, 09:12 PM
beens_angie
Shouldn't the answer be 8^8 divided by 8 factorial?
• Jun 23rd 2006, 09:22 PM
malaygoel
Quote:

Originally Posted by beens_angie
Shouldn't the answer be 8^8 divided by 8 factorial?

No, because arrangement does matter here.You can see it in the third post of this thread.

KeepSmiling
Malay
• Jun 23rd 2006, 09:47 PM
beens_angie
In the third thread I understand that different combinations are given. It does not verify if different arrangements are counted

e.g. 11111112 and
21111111

So I am asking now,....the question specified "combinations", Shouldn't arrangements be excluded?

I agree that my calculation to exclude the arrangements was wrong however I don't think it is simply 8^8 if arrangements are not counted.