Help for combinatorial question?

  • Thread starter erogol
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a "codeword" from the alphabet A={0,1,2,3) is said to be legitimate if it contains even number of zeros. Thus for instance the codeword 31020 is legitimated and 0002 is not. How many n - letter codewords are legitimated ?
 

tiny-tim

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a "codeword" from the alphabet A={0,1,2,3) is said to be legitimate if it contains even number of zeros. Thus for instance the codeword 31020 is legitimated and 0002 is not. How many n - letter codewords are legitimated ?
Hi erogol! :wink:

Show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:
 
14
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i have no sense to solve it i just know answer is 2^(2n-1) + 2^n -1
 

tiny-tim

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i have no sense to solve it i just know answer is 2^(2n-1) + 2^n -1
ok, start by making a sum over all possible (even) numbers of 0s …

the total number of legitimated words is ∑ what ? :smile:
 

HallsofIvy

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So you are not even going to try?
 

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