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Probability Question?

  • 10-04-2016 01:46PM
    #1
    TSMGUY
    Closed Accounts Posts: 855 ✭✭✭


    Okay, this probability question is driving me nuts because I can't understand how the answer in the book makes sense (it's actually a worked example, not a question). Here it is in full:

    HOW MANY DIFFERENT FOUR-DIGIT NUMBERS GREATER THAN 6000 CAN BE FORMED USING THE DIGITS 1,2,3,4,5,6,7 IF I) NO DIGIT CAN BE REPEATED AND II) REPETITIONS ARE ALLOWED

    I got i but ii didn't make sense. I say it's 2 options for the first followed by 7 options followed by 7 options followed by 7 options. So basically 2x7x7x7=686.

    They say it's 2 options followed by 6 options by 6 options by 6 options or 2x6x6x6=432.

    Who's right? If you're allowed repetitions then surely it should be 7x7x7 for the last three digits as you can use any number from 1-7?


Comments

  • #2
    Bowlardo
    Registered Users, Registered Users 2 Posts: 2,390 ✭✭✭Bowlardo


    2(6 or 7) X 6 (123456) X 5 X 4 is what i think the one with not repetions over 6000 I think

    2X7X7X7 would the over 6000 but allowing repetitions


  • #3
    emersyn
    Registered Users, Registered Users 2 Posts: 430 ✭✭emersyn


    Yeah I would've said 2x7x7x7 too


  • #4
    TSMGUY
    Closed Accounts Posts: 855 ✭✭✭TSMGUY


    Bowlardo wrote: »
    2(6 or 7) X 6 (123456) X 5 X 4 is what i think the one with not repetions over 6000 I think

    2X7X7X7 would the over 6000 but allowing repetitions
    The first part is 2x5x4x3 because you've already counted 6 and 7. I definitely don't see any other answer for part ii.
    emersyn wrote: »
    Yeah I would've said 2x7x7x7 too
    THANK YOU! I swear I think I'm going crazy sometimes with how random these answers in the book are. It's so frustrating because I don't even have a teacher to consult with.

    Thanks for the quick responses guys. If anyone else wants to wade in feel free.


  • #5
    skippy1977
    Registered Users, Registered Users 2 Posts: 284 ✭✭skippy1977


    Would agree with Bowlardo for part (i) You have used either 6 or 7 but not both, which leaves you with 6 then 5 then 4.
    2x6x5x4

    Agreed on part (ii) 2x7x7x7


  • #6
    TSMGUY
    Closed Accounts Posts: 855 ✭✭✭TSMGUY


    skippy1977 wrote: »
    Would agree with Bowlardo for part (i) You have used either 6 or 7 but not both, which leaves you with 6 then 5 then 4.
    2x6x5x4

    Agreed on part (ii) 2x7x7x7

    Yeah, looking back on it I'd agree too. It's disgraceful how little effort the authors put into these maths books.


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