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Permutations/combinations with Braille dots

  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)
     
  2. jcsd
  3. Sep 14, 2007 #2

    Gokul43201

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    Try and see how many combinations you can make if you've got only 1, 2, or 3 binary switches (dots) and see if the pattern becomes obvious from them.

    The reason the the two formulas (you missed a 6c0 in the second one) give the same answer is because they are two different ways of counting the same thing. In fact, you've stumbled upon a particular case of a more general identity: [itex] \sum_i C(N,i) = 2^N [/itex]
     
  4. Sep 14, 2007 #3
    im thinking for part b i just go 6c3, 'cause i have 6 choosing 3...and that equals 20...but that seemed to high to me...
     
  5. Sep 14, 2007 #4

    Gokul43201

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    It's actually too low!

    One way to answer that question is to actually go through the procedure and count how many ways you can do it.

    First you have to pick 3 dots and raise them. How many ways to pick 3 dots out of 6?

    Next, you have the freedom to do anything with the remaining 3 dots. How many total configurations can you make out of these three dots?

    These two steps happen in sequence, so what must you do with the numbers you get from each step to find the final answer?
     
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