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Combinatorics of the word RAKSH

  1. Dec 16, 2008 #1
    1. The problem statement, all variables and given/known data
    There is a word given:
    and n slips are provided. A person is free to write any one of the letters (R,A,K,S,H) in each of the slips. Repetition is allowed, i.e. for eg. one such case would be that all the 'n' slips are filled with the letter "R'.

    Then we begin our task:
    First we groups of 1 slip from n
    then groups of 2 slips from n
    then groups of3

    Find the number of such groups formed that contain at least one of each of the letters, i.e. R,A,K,S,H!

    3. The attempt at a solution

    I was told that its a difficult question. Here is what I think:
    it is obvious that groups of 1 to 4 members are useless. since there are 5 letters in RAKSH.

    First I consider those cases in which at least one of each letter is there:
    5 slips have been fixed as RAKSH. and there are remaining n-5 slips , each have 5 options to get filled with.
    so is the answer 5n-5?????
    help me!
  2. jcsd
  3. Dec 16, 2008 #2


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    Re: Combinatorics

    The statement of the problem is not completely clear to me. Is this correct?

    There are n slips of paper. Each slip is printed with a letter. For each slip, the letter has been selected at random, with equal probability, from a list of five letters.

    We now choose k slips at random. What is the probability that each of the five letters occurs at least once?
  4. Dec 16, 2008 #3
    Re: Combinatorics

  5. Dec 17, 2008 #4
    Re: Combinatorics

    I am waiting for some help!!
  6. Dec 17, 2008 #5


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    Science Advisor

    Re: Combinatorics

    Well, if that's the setup, then the total number n of slips doesn't matter. Each of the k slips is equally likely to have any letter on it.

    Can you write down the probability that there are exactly n_R slips with R, n_A slips with A, etc?
  7. Dec 18, 2008 #6
    Re: Combinatorics

    I am studying Permutation an Combination. I havent yet studied probability.
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