Results 1 to 2 of 2

Math Help - fundamental counting principal

  1. #1
    Newbie
    Joined
    Sep 2007
    Posts
    2

    fundamental counting principal

    Could someone please help me, i left my homeowrk to the last minute and it due tomorrow. I don't understand how to complete the question? Even if you could just help with some of the question please. Im in applied math 30, and its the probability unit, about the fundamental counting principle.
    Thx, ahead.

    QUESTION
    Consider the lettersof the word HEXAGON
    a) How many ways can the vowels of the word be arranged using all the vowels?
    b) How many arrangements of the word begain with a vowel?
    c) How many 3-letter "words" can be made if every "word" must have the pattern constant-vowel-constant and if the letters can be repeated?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,802
    Thanks
    691
    Hello, horses7!

    Consider the letters of the word: HEXAGON

    a) How many ways can the vowels of the word be arranged using all the vowels?
    Not sure what this means, but I'll take a guess . . .

    I assume the consonants stay in place, and we are to permute the vowels.

    Then we have: .H _ X _ G _ N

    Then the vowels can be placed in: . 3! \,= \,6 ways.



    b) How many arrangements of the word begin with a vowel?

    There are 7 letters: 3 vowels and 4 consonants.

    There are 3 choices of vowels for the first letter.
    . . Then the other 6 letters can be arranged in 6! = 720 ways.

    Therefore, there are: . 3 \times 720 \:=\:2160 arrangements.



    c) How many 3-letter "words" can be made if every "word" must have the pattern
    consonant-vowel-consonant and if the letters can be repeated?

    For the first letter, there are 4 choices for the consonant.
    For the second letter, there are 3 choices for the vowel.
    For the third letter, there are 4 choices for the consonant.

    Therefore, there are: . 4 \times 3 \times 4 \:=\:48 possible "words".

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Fundamental Counting principle problem
    Posted in the Statistics Forum
    Replies: 2
    Last Post: October 4th 2011, 01:20 PM
  2. More Fundamental Counting
    Posted in the Discrete Math Forum
    Replies: 9
    Last Post: January 12th 2010, 09:27 AM
  3. Fundamental Counting Principles
    Posted in the Discrete Math Forum
    Replies: 10
    Last Post: January 12th 2010, 07:26 AM
  4. [SOLVED] fundamental counting principal.
    Posted in the Statistics Forum
    Replies: 19
    Last Post: May 17th 2009, 11:58 AM
  5. probability and fundamental counting principle?
    Posted in the Statistics Forum
    Replies: 3
    Last Post: March 24th 2008, 05:48 PM

Search Tags


/mathhelpforum @mathhelpforum