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-   -   I think this is a riddle. (https://www.askmehelpdesk.com/showthread.php?t=307123)

  • Jan 22, 2009, 08:43 AM
    latasha johnson
    I think this is a riddle.
    Five people want a photographer to take pictures of every possible group of three of them.
    How many different photos will the photographer need to take?:confused:
  • Jan 22, 2009, 09:03 AM
    Capuchin
    C(5,3)
  • Jan 22, 2009, 12:02 PM
    latasha johnson
    Math, riddle
    Five people want a photographer to take pictures of every possible group of three of them. How many pictures will the photographer need to take?
  • Jan 22, 2009, 12:06 PM
    galactus
    This is not a riddle. It is just wanting to know how many ways you can choose 3 items from 5.

  • Jan 22, 2009, 12:17 PM
    galactus
    Do you know this notation? Factorials ans so forth. Do you have a calculator? Most of them will quickly do it.
  • Jan 22, 2009, 01:13 PM
    Capuchin

    You can do it yourself by going through all the combinations logically too, the answer is not large!
  • Jan 22, 2009, 01:32 PM
    latasha johnson
    Quote:

    Originally Posted by galactus View Post
    This is not a riddle. It is just wanting to know how many ways you can choose 3 items from 5.


    can you show me how to do this step by step?:confused:
  • Jan 22, 2009, 01:36 PM
    latasha johnson
    Quote:

    Originally Posted by Capuchin View Post
    you can do it yourself by going through all the combinations logically too, the answer is not large!

    I still don't understand can you show me step by step please!:confused:
  • Jan 22, 2009, 02:54 PM
    Capuchin

    5! = 5x4x3x2x1, so expanding galactus' answer:

  • Jan 22, 2009, 10:58 PM
    latasha johnson
    Quote:

    Originally Posted by Capuchin View Post
    5! = 5x4x3x2x1, so expanding galactus' answer:


    is the answer a total of 10:confused:
  • Jan 23, 2009, 12:56 AM
    Capuchin

    Yes
  • Jan 23, 2009, 08:51 AM
    latasha johnson
    Quote:

    Originally Posted by Capuchin View Post
    yes

    thank tou:)
  • Jan 23, 2009, 09:45 AM
    Capuchin

    You can also check by working out all the possible combinations yourself:

    The first is obviously

    123

    Then you change the last digit until you run out

    124
    125

    Now you change the 2nd digit and repeat

    134
    135

    And again

    145

    Now you've run out of second digits so change the first:

    234
    235

    Now change the second digit again:

    245

    Now the first again

    345

    Now we've found them all! Count them up and there's 10 :)
  • Jan 23, 2009, 12:03 PM
    latasha johnson
    Quote:

    Originally Posted by latasha johnson View Post
    thank tou:)

    thank you, you have been of great help! :)

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