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anonymous
 5 years ago
Given a boolean function of 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? I had 2^2 + 2^1 + 2^3 + 2^2 = 18 but this includes the repeated terms which I don't know how to minus away to get only the unique number of minterms.
Thanks!
anonymous
 5 years ago
Given a boolean function of 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? I had 2^2 + 2^1 + 2^3 + 2^2 = 18 but this includes the repeated terms which I don't know how to minus away to get only the unique number of minterms. Thanks!

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh, this has to do with the Digital Design course, hmmm, you can use Kmaps to find the unique number of minterms ^_^

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But it has 5 parameters which is quite difficult to use a kmap. So I was thinking it will be better if I could use some combinatorics counting techniques.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Lol, I've never heard of such a way. Which course is this? :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It is a Digital Logic Design course. But I have a feeling that some mathematics counting technique could make this calculation a lot more faster. Expanding the whole boolean expression just to find the number of unique minterms is too time consuming.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh, I haven't taken that :( I took Digital Design and Computer Organization

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So you would also expand out the boolean expression to count the number of unique minterms?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Honestly, I have no idea, the only way I know is using the Kmap.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol... thanks anyway.
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