Help for combinatorial question?

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A legitimate codeword from the alphabet A={0,1,2,3} must contain an even number of zeros, with examples provided to illustrate valid and invalid codewords. The discussion revolves around finding the total number of n-letter legitimate codewords, with one participant stating the answer is 2^(2n-1) + 2^n - 1. Participants encourage sharing attempted solutions to better assist in solving the problem. The conversation highlights the importance of considering all possible even counts of zeros in the codewords. The focus remains on deriving a mathematical solution for the combinatorial question posed.
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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 ?
 
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erogol said:
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:
 
i have no sense to solve it i just know answer is 2^(2n-1) + 2^n -1
 
erogol said:
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:
 
So you are not even going to try?
 
HallsofIvy said:
So you are not even going to try?

Are you talking to me? :biggrin:
 

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