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Braille code probabilities homework

  1. Sep 14, 2007 #1
    Hello everyone im' stuck on this problem.

    It says:
    Each symbol in braille code is represened by a rectangular arrangement of six dots. Given that a least 1 dot of the 6 must be raised, how many symbols can be represented in brail?

    now i saw this posted somewhere else, they got 63. don't know how.

    i got 63 two ways....

    2^6-1 (case where all down)=63

    6c1+6c2+6c3+6c4+6c5+6c6 -1 also =63 (where 6c3 etc is combinations...6 options choose 3)

    can sumone explain how/why that works please.

    also part b...how many combinations have EXACTLY 3 raised

    and how many have an even number of raised dots


    (this is for math and computer science course so not TOO sure which to post it in thanks)
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 14, 2007 #2

    EnumaElish

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    Think of "raised " as "included in a subset," and "flat" as "not included in a subset." Then "all flat" is the empty set. Number of subsets = 2^6 (including empty set).

    How many combinations with exactly 3 raised = number of subsets with 3 elements.

    How many with even raised = number of subsets with 2 or 4 or 6 elements.
     
  4. Sep 14, 2007 #3
    so 2^6 is a? (-1 for original case where non are raised?)

    then for b...2^3...so 8?
    or do i need to figure out how many have more or less than 3 an subtract it?
     
  5. Sep 14, 2007 #4

    Dick

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    The latter. Figure out how many have 0,1 and 2 raised and subtract the total from 2^6.
     
  6. Sep 15, 2007 #5

    HallsofIvy

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    Another way of getting that answer is to argue that each of the 6 dots can be raised or not- a total of 6 "raised" or "not raised" choices. By the counting principal, there would be a total of 26= 64 possiblities. That of course includes "none raised" which is not allowed so there are 64- 1= 63 allowed. That is the same as your sum of binomial coefficients because you are now arguing that it is "number with exactly one dot raised"+ "number with exactly two dots" raised"+ etc. "The number of 6 dots with exactly n raised" is the same as "how many ways can I choose n out of the 6 dots to raise": 6Cn. And, of course, that is 26- 1 because you did not include 6C1= 1. These are binomial coefficients- the coefficients in the expansion of (x+ y)6. The sum of the coefficients is that with x=y= 1: (1+1)6= 26 (and again leaving out the first subtracts 1 from the whole thing).

    As both Enuma Elish and Dick have said- to get "how many have at least three dots raised" Calculate 6C1 and 6C2 (which you may already have done it you calculated that sum of binomial coefficients directly) and subtract from the 63 you already had. (Dick included 6C0 and told you to subtract from 64. Same thing of course.)
     
    Last edited: Sep 15, 2007
  7. Sep 15, 2007 #6

    EnumaElish

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    I have just realized that the OP is an exact replica of the OP in another thread in this forum, created simultaneously with this one.
     
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