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Homework Help: How many combinations?

  1. Sep 29, 2010 #1
    1. The problem statement, all variables and given/known data
    How many 4-letter codes can be formed using the letters A, B, C, D, E, and F? No letter can be used more than once.


    2. Relevant equations
    N/A?



    3. The attempt at a solution
    I really didnt know where to begin . . .
     
  2. jcsd
  3. Sep 29, 2010 #2
    if you can only use a letter once, then just start writing it out on paper. Are you sure a letter can only be used once??? Or once in each position?
     
  4. Sep 29, 2010 #3
    Here is the problem with this question, there is no way to give you a hint without giving you the answer. Seriously.

    If this is assigned from a textbook, then the answer is there. The hint is that it is not combinations, that term is not the math term for it.
     
  5. Sep 29, 2010 #4
    Im not sure how this is so easy for you. I assume that the question means that no lettter can be used more than once within one of the 4-letter codes. If this is the case, then there are way to many codes to write by hand.

    This problem came from a review worksheet. If this is not a combinations problem, then what is it?
     
  6. Sep 29, 2010 #5
    Well, think about it. You've got 5 letters, right? And you know that you can't use each letter more than once. So what does that imply each time you've chosen a letter?
     
  7. Sep 29, 2010 #6
    This implies that after each time I have chosen a letter, there is one less to choose from.

    I tried to use n! to find an answer by doing 6x5x4x3x2x1, but the resulting answer was way to large.
     
  8. Sep 29, 2010 #7
    Okay, so you know that the available choices diminish by 1 each time you choose a letter. Only other hint I can think of without giving the answer away is to consider if the order the letters appear in is important. IE: Is, say, FBACE the same code as ECABF?
     
  9. Sep 29, 2010 #8
    I would think the order is important, but how does that help me?
     
  10. Sep 29, 2010 #9
    Well, check your textbook. What are you dealing with when order is important?
     
  11. Sep 29, 2010 #10
    I dont have a textbook, but im thinking permutations. I dont really know how they differ functionally from a combination though.
     
  12. Sep 29, 2010 #11
    Sounds good. Permutations are employed when the order in which the objects are chosen matters, such as when ABCDE (in this case) is not the same as EDCBA. If the order of the letters didn't matter (Such as if I asked you to tell me how many ways you can throw 5 random letters together rather than asking you for 5-letter codes) you'd use the formula for combinations instead.
     
  13. Oct 3, 2010 #12
    Ok, so I should use n!/(n-r)!

    so, 6!/(6-4)! = 360?
     
  14. Oct 3, 2010 #13
    The formula you want is one that is used for combinations, i THINK. You want 6 choose 4 or in other words, how many ways can you choose 4 from 6. The formula for n choose r is [tex]\frac{n!}{(n-r)!r!} \Rightarrow \frac{6!}{(6-4)!4!}[/tex]
     
    Last edited: Oct 3, 2010
  15. Oct 3, 2010 #14
    Originally, I had thought this too, however, I think the order matters here, so the permutation formula must be used.
     
  16. Oct 3, 2010 #15
    ok sorry. I might do [tex]6 \times 5 \times 4 \times 3[/tex]
     
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