# Number of n-length "words"

1. Oct 24, 2016

### Lilia

1. The problem statement, all variables and given/known data
Given an alphabet of {0,1,2}, how many "words" of length n are there that contain even 0s?

2. Relevant equations
Choose 2k 0s from n - C(n,2k), k=0,n/2

3. The attempt at a solution
I tried to solve this for n=4 and n=5. For n=4 I got 12 (or, if 0000 is also counted then 13), for n=5 - 30. But I can't figure out the formula

2. Oct 24, 2016

### arpon

First, consider this problem : How many words of length $n$ contains $2k$ $0$s?
There are $^nC_{2k}$ ways to choose the $2k$ places for $2k$ $0$s. After setting $0$s, we have $n-2k$ places to be filled up by $1$s and $2$s. We can use as many $1$s and $2$s as we like. So there are $2^{n-2k}$ ways to fill the rest $n-2k$ places by $1$s and $2$s.
Therefore, there are $^nC_{2k}\cdot 2^{n-2k}$ words of length $n$ that contain $2k$ 0s.
Can you figure out the formula now?
[Hints: Apply binomial theorem]

Last edited: Oct 24, 2016
3. Oct 24, 2016

### Lilia

I know what is the binomial theorem but I don't know how to transform this formula to get the binomial form of it

4. Oct 24, 2016

### haruspex

Consider the expansion of (1+x)n. Your problem is that you get both odd and even powers of x. How could you add another expansion to make only the odd powers disappear?