Results 1 to 7 of 7

Math Help - I Need Help With Factors

  1. #1
    Newbie
    Joined
    Dec 2006
    Posts
    3

    I Need Help With Factors

    So I Know Nothing About Math And I Need To Get This Done. Here Is The Problem

    I Have A Footprint Of A House. And There Are 7 Options That Could Change That Footprint. So How Many Footprints Can There Be Total?

    1 Footprint
    7 Options

    You Cannot Use The Same Option. So If I Used Option A And Added Option B, When It Comes To Figure Out The Options On B I Cant Use A. Make Sense? I Think The Answer Is 148 But Im Not Sure. Please Help Anyone.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,547
    Thanks
    539
    Hello, HVRIDERZ!

    I know nothing about a footprint of a house . . . what is it?


    I Have A Footprint Of A House.
    And There Are 7 Options That Could Change That Footprint.
    So How Many Footprints Can There Be Total?

    For each option, there are two choices: apply it or don't apply it.

    Hence, there are: 2^7 = 128 possible options
    . . from "use none of them" to "use all of them".

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member OReilly's Avatar
    Joined
    Mar 2006
    Posts
    340
    Quote Originally Posted by HVRIDERZ View Post
    So I Know Nothing About Math And I Need To Get This Done. Here Is The Problem

    I Have A Footprint Of A House. And There Are 7 Options That Could Change That Footprint. So How Many Footprints Can There Be Total?

    1 Footprint
    7 Options

    You Cannot Use The Same Option. So If I Used Option A And Added Option B, When It Comes To Figure Out The Options On B I Cant Use A. Make Sense? I Think The Answer Is 148 But Im Not Sure. Please Help Anyone.

    If I understand it correctly you want find every possible different combinations of options to apply on footprint.

    So, you have:
    7/1 = 7
    7*6/1*2 = 21
    7*6*5/1*2*3 = 35
    7*6*5*4/1*2*3*4 = 35
    7*6*5*4*3/1*2*3*4*5 = 21
    7*6*5*4*3*2/1*2*3*4*5*6 = 7
    7*6*5*4*3*2*1/1*2*3*4*5*6*7 = 1

    Add all those results:
    7+21+35+35+21+7+1=127 total options.

    Total 128 various footprints (127 with different options and one footprint without applying any option).
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Quick's Avatar
    Joined
    May 2006
    From
    New England
    Posts
    1,024
    Quote Originally Posted by Soroban View Post
    Hello, HVRIDERZ!

    I know nothing about a "footprint of a house" . . . what is it?



    For each option, there are two choices: apply it or don't apply it.

    Hence, there are: 2^7 = 128 possible results,
    . . from "use none of them" to "use all of them".

    Deja vu
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Dec 2006
    Posts
    3
    ok a footprint of a house is the perimeter of a house. so take the house u live in and measure all the outside walls, then draw it on a piece of paper and you will have a footprint.

    so one factor i forgot to mention, 2 options cant be used together. just some info: the option is a 2' garage extention and a 4' garage extention. so clearly you can only have one or the other. what would be the formula to finding that out.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member OReilly's Avatar
    Joined
    Mar 2006
    Posts
    340
    Quote Originally Posted by HVRIDERZ View Post
    ok a footprint of a house is the perimeter of a house. so take the house u live in and measure all the outside walls, then draw it on a piece of paper and you will have a footprint.

    so one factor i forgot to mention, 2 options cant be used together. just some info: the option is a 2' garage extention and a 4' garage extention. so clearly you can only have one or the other. what would be the formula to finding that out.
    Subtract from number 128 possible combinatios of that two options that can't be together with other options.

    Let's say you have options A,B,C,D,E,F,G so that F,G can't be together in any combination.

    We have already counted total number of options (and with no options) which is 128.

    So, we must find total number of combinations where F,G are together.
    That is total of 32 combinations where F,G are together.

    That is 128 - 32 = 96 of total options.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Dec 2006
    Posts
    3
    can you tell me how you figured out 32? the only thing i came up with was 2^5 and that equals 23. but what is the reasoning for that?

    i drew it up and came up with 96, now if i use the same thing as you did of 2^5 and try to simplify the same situation so now i have 4 options or in this case now known as (G,F,E,D,) so this is the deal. G and F can never be together. same goes for E and D. however G can be with E or D but not the 2 together. hopefully all that makes sense.im trying to get a formula to do it so i can proceed with my work. thanks
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. factors
    Posted in the Number Theory Forum
    Replies: 3
    Last Post: April 7th 2010, 07:26 AM
  2. Replies: 1
    Last Post: December 7th 2009, 10:42 AM
  3. sum of factors
    Posted in the Number Theory Forum
    Replies: 3
    Last Post: June 25th 2009, 06:23 AM
  4. Factors of 2310 and Factors of 1365
    Posted in the Algebra Forum
    Replies: 1
    Last Post: October 7th 2008, 06:56 PM
  5. Factors
    Posted in the Algebra Forum
    Replies: 1
    Last Post: December 4th 2007, 04:37 AM

Search Tags


/mathhelpforum @mathhelpforum