Results 1 to 7 of 7

Math Help - Probability

  1. #1
    Junior Member
    Joined
    Nov 2007
    Posts
    27

    Wink Probability

    I have got stuck on this question, is anyone able to help me please (any help is welcome):

    Given an alphabet of 26 letters, how many sets of initials can be formed if every person has one surname and

    i) exactly 2 given names

    ii) at most 2 given names

    iii) at most 3 given names

    iv) The conclusion is that in a town of 20000 inhabitants then either some people have the same set of initials or at least x have more than 3 initials. How large is x?

    Many thanks to anyone who has any advice.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Mar 2008
    Posts
    5
    Here is what I'd do.....

    (i) There are 26 possible letters, each set of initials would have 3 letters in it, order doesnt matter, i.e. ABC does not equal ACB. So type in 26P3 on your calculator (P=permutation).

    (ii) This time each set of initials can have 2 or 3 values, so calculate 26P3 + 26P2.

    (iii) 26P4 + 26P3 + 26P2.

    (iv) x = solution for (ii) - 20,000

    Hope this helps...
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Nov 2007
    Posts
    27
    roegamma,

    thanks for ur help, by permutation (P) do u mean type 26^3 into the calculator.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Mar 2008
    Posts
    5
    No, you should have a P button somewhere on your calculator. Could be signified by nPr.

    http://www.maplin.co.uk/images/Full/A92FQ.jpg

    Thats pretty much my calculator. For me to do 26P4 I'd have to type 26, shift, x, 4.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Nov 2007
    Posts
    27
    roegamma

    oh yeah i found it thanks. one last question, for part (iv) should it be 20000 - (26P3 + 26P2) otherwise if i did (26P3 + 26P2) - 20000 then i would get a negative number?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Mar 2008
    Posts
    5
    Yeah, sorry thats my fault. It should be 2000 - answer to part (ii).
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1574
    Awards
    1
    Quote Originally Posted by smiler View Post
    Given an alphabet of 26 letters, how many sets of initials can be formed if every person has one surname and
    i) exactly 2 given names
    ii) at most 2 given names
    iii) at most 3 given names
    iv) The conclusion is that in a town of 20000 inhabitants then either some people have the same set of initials or at least x have more than 3 initials. How large is x?
    There is nothing about the statement of this problem to lead one to conclude that a person has three different initials. I know someone named August Aaron Adams. His initials are AAA.

    Quote Originally Posted by roe_gamma View Post
    Here is what I'd do.....
    (i) There are 26 possible letters, each set of initials would have 3 letters in it, order doesnt matter, i.e. ABC does not equal ACB. So type in 26P3 on your calculator (P=permutation).
    (ii) This time each set of initials can have 2 or 3 values, so calculate 26P3 + 26P2.
    (iii) 26P4 + 26P3 + 26P2.
    (iv) x = solution for (ii) - 20,000
    The above answers assume that the initials are all distinct in a name. That is an unwarranted conclusion.
    Thus the correct answers:
    i) {26}^3.

    ii) 26 + {26}^2+{26}^3 [no given name, one given or two] (Now you may want to assume that everyone has at least one given name.)

    iii) is like #2 26 + {26}^2+{26}^3 + {26}^4

    To see (iv) note that from (iii) 26 + {26}^2+{26}^3  = 18278 (or if everyone must have a given name then {26}^2 + {26}^3 = 18252). So in any case in a town of 20000 by the pigeonhole principle some people must have the same initials.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 10
    Last Post: January 21st 2011, 11:47 AM
  2. Replies: 0
    Last Post: December 6th 2010, 04:57 PM
  3. Replies: 3
    Last Post: May 29th 2010, 07:29 AM
  4. Replies: 1
    Last Post: February 18th 2010, 01:54 AM
  5. Replies: 3
    Last Post: December 15th 2009, 06:30 AM

Search Tags


/mathhelpforum @mathhelpforum