Results 1 to 4 of 4

Math Help - Combinatorics

  1. #1
    Junior Member
    Joined
    Jul 2009
    Posts
    26

    Question Combinatorics

    hi

    i don't know where else to post this but i need some help on a textbook question i was given about combinatorics.

    there is a game that involves a murderer, a mansion, and the weapon used to commit the murder. there are 5 characters in all, 7 rooms in the mansion, and 3 weapons.

    the question is.. how many kinds of guesses can i make from choosing one character, one mansion room, and one weapon?

    what kind of formula do i use for this, nPr or nCr? or can i use the fundamental counting principle?

    any help?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jul 2009
    Posts
    593
    Thanks
    4
    What you want is n_1!*n_2!*n_3!, where n_1 is the Character, n_2 is the room in the mansion, and n_3 is the weapon.

    For example: Lets say Professor Plum we know is the murderer. He could have done it in any of the 7 rooms correct? So there are (at the moment) 7 possible scenarios for Plum to have committed the murder. But we also know he can use 3 weapons in each of the 7 scenarios: so thats three weapons in Room 1, three weapons in Room 2, three weapons in Room 3, etc. to three weapons in room 7 - for a total of 21 possible scenarios that Professor Plum (Character), could have gone somewhere (Room) and used a weapon (Weapon) to murder someone. But Plum isn't the only character. We have 5.

    Thus you would use the basic methods of counting to find out the total possible scenarios that can play out here.

    As an aside, remember what permutations and combinations are: Permutations are just an arrangement of n objects, and combinations are the same arragement with uniqueness factored in.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jul 2009
    Posts
    26
    Quote Originally Posted by ANDS! View Post
    What you want is n_1!*n_2!*n_3!, where n_1 is the Character, n_2 is the room in the mansion, and n_3 is the weapon.

    For example: Lets say Professor Plum we know is the murderer. He could have done it in any of the 7 rooms correct? So there are (at the moment) 7 possible scenarios for Plum to have committed the murder. But we also know he can use 3 weapons in each of the 7 scenarios: so thats three weapons in Room 1, three weapons in Room 2, three weapons in Room 3, etc. to three weapons in room 7 - for a total of 21 possible scenarios that Professor Plum (Character), could have gone somewhere (Room) and used a weapon (Weapon) to murder someone. But Plum isn't the only character. We have 5.

    Thus you would use the basic methods of counting to find out the total possible scenarios that can play out here.

    As an aside, remember what permutations and combinations are: Permutations are just an arrangement of n objects, and combinations are the same arragement with uniqueness factored in.
    Ohhh.. so to work it out, it would be 5! x 7! x 3! ??
    thanks
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,711
    Thanks
    1640
    Awards
    1
    Quote Originally Posted by oryxncrake View Post
    there is a game that involves a murderer, a mansion, and the weapon used to commit the murder. there are 5 characters in all, 7 rooms in the mansion, and 3 weapons.
    the question is.. how many kinds of guesses can i make from choosing one character, one mansion room, and one weapon?

    what kind of formula do i use for this, nPr or nCr? or can i use the fundamental counting principle?
    use the fundamental counting principle: (5)(7)(3).
    The question is about content not order.
    Therefore no factorials are necessary.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Combinatorics.
    Posted in the Discrete Math Forum
    Replies: 16
    Last Post: July 20th 2010, 02:29 AM
  2. Combinatorics
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: June 18th 2010, 08:14 PM
  3. Combinatorics
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: June 3rd 2010, 05:24 PM
  4. combinatorics
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: May 1st 2010, 10:53 PM
  5. Combinatorics
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: October 10th 2009, 06:03 AM

Search Tags


/mathhelpforum @mathhelpforum