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Thread: I have 3 variables...

  1. #17
    Pre-Cambrian nibbler's Avatar
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    Re: I have 3 variables...

    Quote Originally Posted by Phage View Post
    28...
    3 days! School been **** since monday, had a bit of lovely maths then but none since. I'm habing withdrawal symptoms
    And yes, it's 1000 solutions or 720, depending on which way you look at it.

    Edit: Whut? Why is the word "shut" starred out?
    Edit: Oh I see what happened there
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    I R Toff Pandi! TAKTAK's Avatar
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    Re: I have 3 variables...

    2 hours 11 minutes for me...

    No nostalgia needed...
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    Re: I have 3 variables...

    That'll teach me to post without trying the common sense method

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  5. #20
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    Re: I have 3 variables...

    Quote Originally Posted by TAKTAK View Post
    2 hours 11 minutes for me...

    No nostalgia needed...
    Enjoy maths while you can.

    College is just beer, pizzas and all-niter assignments the day before they're due. It sounds a lot better on paper than it actually is trust me on this. Come to think of it I don't even like beer or assignments...

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    Re: I have 3 variables...

    Quote Originally Posted by AD-15 View Post
    In that case, it would be 999 (1000 is a four digit number ).

    However, the OP stated that the 3 numbers are different - so you couldn't have (for example): 111, 112, 099, 444, 696 etc...).
    If they are different, then the calculation is 10!-7!
    ! is the symbol for factorial IIRC. It's best illustrated with an example.
    5! is 5*4*3*2*1
    Disclaimer: it's been 14 years since I did maths and I have never used this kind of Maths since so I may be way off!
    "In a perfect world... spammers would get caught, go to jail, and share a cell with many men who have enlarged their penises, taken Viagra and are looking for a new relationship."

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    Re: I have 3 variables...

    The calculation is most certainly not 10!-7!, it's simply 10^3 as scaryjim has already told us.
    To find the number of combinations you take the number of possible figures (10 in this case) and then the number of positions (3 in this case), let them be n and k respectively then do n^k. It gets more complex if you want different numbers and that's when you need to use common sense like snooty did and do 10*9*8 = 720.
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    Not a good person scaryjim's Avatar
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    Re: I have 3 variables...

    Indeed, it's not 10! - 7!, it's 10! / 7!. Which works out as 10*9*8, as nibbler says.

    I learned everything I know about probabilities and combination from a GCSE Maths project on Poker: so, not very much at all then

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    Re: I have 3 variables...

    Quote Originally Posted by scaryjim View Post
    Indeed, it's not 10! - 7!, it's 10! / 7!. Which works out as 10*9*8, as nibbler says.

    I learned everything I know about probabilities and combination from a GCSE Maths project on Poker: so, not very much at all then
    My mistake - it was 10!/7!
    "In a perfect world... spammers would get caught, go to jail, and share a cell with many men who have enlarged their penises, taken Viagra and are looking for a new relationship."

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