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How to find all different combinations of six numbers



 
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jgreer
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PostPosted: Thu Feb 15, 2007 7:31 am    Post subject: How to find all different combinations of six numbers
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This is not a homework assignment, just something I am having trouble figuring out and thought someone may be able to help.

How many different combinations of six numbers are there using all of the numbers between 1 and 45 Question

i.e. 1,2,3,4,5,6 - 1,2,3,4,5,7 - 1,2,3,4,5,45 - - - 40,41,42,43,44,45

I have tried to figure this out a few times but constantly get confused or think i have left some out.

note: The combinations do not include those that already have the same numbers i.e. If you had 1,2,3,4,5,6 you would not need 6,5,4,3,2,1 as a combination as all six numbers are already counted as a combination.

Please leave your answers to this problem if you choose to try to figure it out as it will be interesting to see what everyone comes up with.

Cheers
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jgreer
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PostPosted: Thu Feb 15, 2007 8:15 am    Post subject:
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Below is a further example of the combinations so nothing is forgotten

1,2,3,4,5,6
1,2,3,4,5,7
1,2,3,4,5,8
1,2,3,4,5,45

1,3,4,5,6,7
1,3,4,5,6,8
1,3,4,5,6,45

1,2,4,5,6,7
1,2,4,5,6,8
1,2,4,5,6,45

1,2,3,5,6,7
1,2,3,5,6,8
1,2,3,5,6,45

1,2,3,4,6,7
1,2,3,4,6,8
1,2,3,4,6,45

1,2,3,4,5,7
1,2,3,4,5,8
1,2,3,4,5,45

2,3,4,5,6,7
2,3,4,5,6,8
2,3,4,5,6,45

and so on

40,41,42,43,44,45 will be the last combination.
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chris
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PostPosted: Thu Feb 15, 2007 12:55 pm    Post subject:
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How many combination can you build if you fix the first 5 numbers? Right - 40. Because you have 45-5=40 numbers to choose from and make it the sixth number.

How many ways do you have to build the combinations of those 5 numbers? Notice that now the problem is the same, but we have one less, i.e. 5 numbers instead of 6 to build our combinations. Apply the same thought as above: suppose the first 4 are fixed, how many combinations do you have for the 5th?

And so on and so on...till you have only one number and 45 choices.

This gives you:

45 x (45-1) x (45-2) x (45-3) x (45-4) x (45-5)
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Kriffo
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PostPosted: Mon Mar 05, 2007 2:28 pm    Post subject:
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Wouldnt that give you too many since he says that 1,2,3,4,5,6 and 6,5,4,3,2,1 are the same.

I would guess you need to do: ` (45!)/(6!(45-6)!)`
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chris
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PostPosted: Sun Mar 11, 2007 10:28 am    Post subject:
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Kriffo wrote:
Open quoteWouldnt that give you too many since he says that 1,2,3,4,5,6 and 6,5,4,3,2,1 are the same. Close quote


Right, I missed that, or I wanted him to think about it, not give the the whole answer - not sure which of the two. Mr. Green
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rovent
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PostPosted: Sun Mar 25, 2007 10:18 pm    Post subject:
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Well, if you mean how many different combinations of numbers you can compose of 1 to 45 , WITH ORDER, (meaning 1,2,3,4,5,6 is different than 6,5,4,3,2,1) then chris is right. if you mean w/o order (which i think is what you meant) then its C(45,6) = what chris said /6! = what kriffo said (6! is for the internal order of each option of numbers which is the number of permutations over {1..6} = 6!).
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