Hi erogol!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 ?
ok, start by making a sum over all possible (even) numbers of 0s …i have no sense to solve it i just know answer is 2^(2n-1) + 2^n -1
Are you talking to me?So you are not even going to try?