Results 1 to 3 of 3

Math Help - Please help me with this finite question

  1. #1
    Newbie
    Joined
    Sep 2009
    Posts
    6

    Question Please help me with this finite question

    Assume that the representative will continue to make
    telephone calls until she makes
    2 sales, gets 5 tentative commitments, or fails to make a sale.
    Her log for the day consists of a list of calls, with sale, tentative
    commitment, or no sale noted for each.



    How many possible logs are there?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    Hello jakeiu

    Wecome to Math Help Forum!
    Quote Originally Posted by jakeiu View Post
    Assume that the representative will continue to make
    telephone calls until she makes
    2 sales, gets 5 tentative commitments, or fails to make a sale.
    Her log for the day consists of a list of calls, with sale, tentative
    commitment, or no sale noted for each.



    How many possible logs are there?
    Consider first those logs that end because she makes 2 sales (S). Then she must get no 'fails' (F) (or the log will end immediately), and she can get from 0 to 4 'tentatives' (T). Let n(S), n(T), n(F) denote the number of sales, ... etc.

    So, when the log ends with n(S) = 2, n(F) = 0 and n(T) = 0, 1, ..., 4

    With n(T) = 0, there is just one possible log: S, S.

    With n(T) = 1, there are two possible logs: T, S, S and S, T, S.

    With n(T) = 2, the number of logs is the number of ways of choosing which position 1, 2 or 3 the S will come, which is 3,

    With n(T) = 3, the number of logs is 4.

    With n(T) = 4, the number of logs is 5.

    Total for n(S) = 2 is 1 + 2 + 3 + 4 + 5 = 15


    If n(T) = 5, then once again n(F) = 0, and n(S) = 0 or 1.

    If n(S) = 0, there is just one log.

    If n(S) = 1, there are 5 positions in which the S could come.

    So the total for n(T) = 5 is 1 + 5 = 6


    If n(F) = 1, then the F must come at the end of the sequence, with n(S) = 0, 1 and n(T) = 0, 1, ..., 4

    If n(S) = 0, there are 5 possible logs, corresponding to each of the five values of n(T).

    If n(S) = 1, there are 1 + 2 + 3 + 4 + 5 = 15 possible logs (each of these individual numbers being the number of ways of positioning the S with 0, 1, ..., 4 T's)

    So the total for n(F) = 1 is 5 + 15 = 20


    So I reckon the overall total is 15 + 6 + 20 = 41 logs.

    Grandad
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2009
    Posts
    6
    Thank you so much for that answer!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Distinct Bases for Finite Vector Spaces over Finite Fields
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: May 29th 2011, 12:31 PM
  2. Finite sum question
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 26th 2011, 05:04 AM
  3. finite group question
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 16th 2010, 08:29 PM
  4. Another finite question I need help with please
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: September 11th 2009, 08:17 AM
  5. Finite function question
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: September 25th 2008, 03:27 PM

Search Tags


/mathhelpforum @mathhelpforum