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Old June 25th, 2008, 03:19 PM   #1
ElDeeder
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Default Help with math

For a class a friend is taking. License plates.

2 number and 4 letters.

3 numbers and 3 letters.

How to figure out how many combinations are possible.



Thanks,
ElDeeDer

Last edited by ElDeeder; June 25th, 2008 at 03:24 PM.
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Old June 25th, 2008, 03:20 PM   #2
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Bigcunt, here is your chance!!!
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Old June 25th, 2008, 03:20 PM   #3
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BigCountry is your best bet on getting it correct
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Old June 25th, 2008, 03:20 PM   #4
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I would ask bigcountry, but my guess is 4/4=4
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Old June 25th, 2008, 03:26 PM   #5
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2 Number and 4 Letter combo:
10 x 10 x 26 x 26 x 26 x 26*

3 Num 3 Letter:
10 x 10 x 10 x 26 x 26 x 26*

*If letters and numbers can be used more than once.

imho.... Go with BigCountry in 4/4=4
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Old June 25th, 2008, 03:27 PM   #6
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45,697,600

17,576,000
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Old June 25th, 2008, 03:28 PM   #7
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Is this friend learning how to press license plates in the pen?
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Old June 25th, 2008, 03:30 PM   #8
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Quote:
Originally Posted by Hodur308 View Post
Is this friend learning how to press license plates in the pen?
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Old June 25th, 2008, 03:30 PM   #9
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Quote:
Originally Posted by Hodur308 View Post
Is this friend learning how to press license plates in the pen?
they have excell in the pen??? aside from the surpprise butsecks jail is lookin pretty good...free cable, 3 meals a day, medical care, excell spreadsheet capability....
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Old June 25th, 2008, 03:31 PM   #10
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Calculating combinations and permutations
combinations (nCk)
Click in an unoccupied cell, for example C4. Then type =combin(your n-value, your k-value), i.e. "10 choose 6" would be =combin(10,6). The number of combinations will appear in the C4 box. It should be 210.
permutations (nPk)
Click in an unoccupied cell, for example C5. Then type =permut(your n-value, your k-value), i.e. 9 P 5 would be =permut(9,5). The number of permutations will appear in the C5 box. It should be 15120.


excel for the win.
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Old June 25th, 2008, 03:37 PM   #11
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the answer is = ammount of posts you have - the IQ of bigcountry + the number of cars broken into by AjHall / how many times he got caught * the number of permits he had
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Old June 25th, 2008, 03:41 PM   #12
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In general, the number of ways to choose k items from n is:

n!/(k!(n-k)!)

thus, 2 of 10 + 4 of 27 = answer 1
3 of 10 + 3 of 27 = answer 2
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Old June 25th, 2008, 03:45 PM   #13
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Lothos,

When I put =combin(10,26) in the box i get #num!

10 being 0-9
26 being 26 letters in the alphabet.

Am i not following something or am i missing something.

Thanks for the help.
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Old June 25th, 2008, 03:46 PM   #14
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=combin(2,10)+ =combin(4,26) = answer 1
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Old June 25th, 2008, 03:48 PM   #15
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Gotcha, thank you.

Whenver I put =combin(4,26) i still get #NUM!
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Old June 25th, 2008, 03:50 PM   #16
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expand the cell wider or check the cell format settings.
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Old June 25th, 2008, 03:58 PM   #17
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Quote:
Originally Posted by Lothos View Post
=combin(2,10)+ =combin(4,26) = answer 1
The second parameter can't be larger than the first, otherwise you get an error.
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Old June 25th, 2008, 03:59 PM   #18
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oh yeah, forgot the the first number is the set and the second is the number of items from the set.
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Old June 25th, 2008, 05:49 PM   #19
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You forgot to factor in the plates/combinations that the state won't allow (like ASS as the three letters, etc).
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Old June 25th, 2008, 07:05 PM   #20
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this thread got intelligent....we need a farm_boy comment to bring this back to pub standards.
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