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Thread: How do I minus away repeated combinations?

  1. #1
    Oct 2010

    How do I minus away repeated combinations?

    My question has a little relation to Digital Logic design. But what I want to ask is how I could calculate through the use of combinatorics counting techniques.

    Given a boolean function F(A,B,C,D,E) = CD' + ABD'E' + DE + A'BE

    How do I find the number of unique minterms through the use of combinatorics? First, I listed out a table of bit string to visualise:
    A B C D E
    x x 1 0 x
    1 1 x 0 0
    x x x 1 1
    0 1 x x 1
    "x" are the possibility of being a 1 or 0.

    Then I had 2^2 + 2^1 + 2^3 + 2^2 = 18, which is the permutation of the terms. But this includes the repeated terms which I don't know how to minus away to get only the unique number of minterms.

    How do I find out the number of repeated minterms and minus from 18?

    I am sorry if I've posted this in a wrong section.

    Last edited by xEnOn; Apr 22nd 2011 at 04:46 AM.
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  2. #2
    Jun 2010
    It may be hard for us to answer this question because of the non-mathematical concepts involved. I know a bit about combination calculations, but I have no idea what a minterm is, what the prime symbols mean in the Boolean function, etc.

    If I am to guess about my understanding here, I'm thinking that the repeats you're trying to subtract out are things like the first row and second row being 11100. If that's the case, I know no general combinatoric way of treating this. I would observe that there is only one combination which is possibly repeated between the first two rows. The third row cannot repeat any other row than the last one, and there are two possible ways these could happen. The only other possible way of finding repetition is with the first and last, and there are two possibilities for such repetition. This totals to five repetitions.

    However, there are no possible ways of gaining a repetition among all of the rows at the same time. I don't know if that matters here.
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