Help for combinatorial question?

[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 ?

 

[tiny-tim]

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!

Show us what you've tried, and where you're stuck, and then we'll know how to help!

 

[erogol]

i have no sense to solve it i just know answer is 2^(2n-1) + 2^n -1

 

[tiny-tim]

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 ?

 

[HallsofIvy]

So you are not even going to try?

 

[tiny-tim]

So you are not even going to try?

Are you talking to me?