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Problem Solving with Probability

  1. Apr 2, 2006 #1
    Hi, I just need people to check my work.
    1.
    a) Determine the number of ways to arrange the letters in the word BOOKKEEPER.
    I did [tex]\frac{10!}{2!2!3!}[/tex]
    which is 151200.

    b) Determine the number of ways to arrange the letters in the word BOOKKEEPER if you select at most 3.
    I did [tex]\frac{10P_3}{2!2!3!}[/tex]
    which is 30.


    c) Determine the number of ways to arrange the letters in the word BOOKKEEPER if you select only 5.
    I did [tex]\frac{10C_5}{2!2!3!}[/tex]
    which is 10.5.
    I know c) can't be a decimel.
     
    Last edited: Apr 3, 2006
  2. jcsd
  3. Apr 2, 2006 #2
    for part a),
    How many distinct letters make up the word BOOKEEPER?
    What is the repetition number of each letter?
     
  4. Apr 2, 2006 #3
    You have the following letters:

    1 B
    2 Os
    1 K
    3 Es
    1 R

    So you can rearrange the Os and the Es. There are Two ways to rearrange the Os and Six ways to rearrange the Es. and 2 * 6 is 12. Is that not right?

    Do you have to spell Bookeeper or can they be in any order?
     
  5. Apr 2, 2006 #4
    There are 6 distinct letters make up the word BOOKEEPER and the repetition number of each letter are lised below.
    B O K E P R
    1 2 1 3 1 1

    I don't see how this helps me answer the question though.

    EDIT:Oh, I made a mistake there, I don't know why I had another 2!. I just edited my work, is it alright now?

    Thanks :)
     
    Last edited: Apr 2, 2006
  6. Apr 2, 2006 #5
    how did you come up with 10! in the numerator?

    (how many letters are in the word BOOKEEPER ?)
     
    Last edited: Apr 2, 2006
  7. Apr 2, 2006 #6
    The question asks to count the number of distinct arrangments of the word BOOKEEPER.
    So for example one such arrangement is OEBEPKORE and another is ROOKEEPEB.
    Notice if we swap the positions of the O's in the last example we obtain the same arrangement as before. This is because the O's are considered indestinguishable, we can not tell one O from the other. Same thing with the E's.
     
  8. Apr 3, 2006 #7
    Oh, the reason why I had 10! and another 2! was because it was suppose to spell BOOKKEEPER with 2 K's. I was addled with the lettering of the word.

    EDIT:
    I just changed it, is c) correct?
     
  9. Apr 3, 2006 #8
    Ok, adding that extra K makes your original answer to part (a) correct. As its written now though, you forgot to edit back in the extra factor of 2! in the denominator. Obviously I didn't know how to spell BOOKKEEPER either :biggrin:

    I would do (c) first, and then use similar methods to solve (b).

    Obviously, as you stated, part (c) is incorrect as the answer can not be rational.
    suppose we want to calculate the number of arrangements of BOOKE, then using the methods from part (a), we get that this equals,
    [tex] \frac{5!}{2!} = 60[/tex]

    now lets do the same for BOOEE, we get
    [tex] \frac{5!}{2!2!} = 30 [/tex]

    To get all possible arrangements of size 5 from the word BOOKKEEPER, we must perform this calculation for each of the possible ways of selecting 5 letters from BOOKKEEPER, and add our results. Also keep in mind that BOOKE (selecting the first K and third E), is the same as BOOKE (selecting the second K and first E) because the K's and E's are indestinguishable.
     
    Last edited: Apr 3, 2006
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