Let $m$ be a positive integer and let $A$, respectively $B$, be two alphabets with $m$, respectively $2m$ letters. Let also $n$ be an even integer which is at least $2m$. Let $a_n$ be the number of words  of length $n$, formed with letters from $A$, in which appear all the letters from $A$, each an even number of times. Let $b_n$ be the number of words of length $n$, formed with letters from $B$, in which appear all the letters from $B$, each an odd number of times. Compute $\frac{b_n}{a_n}$.
To solve the problem, we need to compute the ratio \(\frac{b_n}{a_n}\), where \(a_n\) is the number of words of length \(n\) formed with letters from \(A\) (each letter appearing an even number of times), and \(b_n\) is the number of words of length \(n\) formed with letters from \(B\) (each letter appearing an odd number of times).

1. **Understanding the alphabets and words**:
   - Alphabet \(A\) has \(m\) letters.
   - Alphabet \(B\) has \(2m\) letters.
   - \(n\) is an even integer and \(n \geq 2m\).

2. **Counting \(a_n\)**:
   - Each letter from \(A\) must appear an even number of times in a word of length \(n\).
   - Since \(n\) is even, we can partition \(n\) into \(m\) pairs of even counts.
   - The number of such words is given by the multinomial coefficient:
     \[
     a_n = \binom{n}{2, 2, \ldots, 2} = \frac{n!}{(2!)^m}
     \]

3. **Counting \(b_n\)**:
   - Each letter from \(B\) must appear an odd number of times in a word of length \(n\).
   - Since \(n\) is even, we can partition \(n\) into \(2m\) odd counts.
   - The number of such words is given by the multinomial coefficient:
     \[
     b_n = \binom{n}{1, 1, \ldots, 1} = \frac{n!}{1!^{2m}} = n!
     \]

4. **Bijection between \(a_n\) and \(b_n\)**:
   - Consider a word from \(a_n\). Each letter from \(A\) can be replaced by a pair of letters from \(B\).
   - For each letter in \(A\), there are \(2\) corresponding letters in \(B\).
   - This replacement ensures that each letter from \(B\) appears an odd number of times.

5. **Applying the lemma**:
   - The lemma states that the number of binary strings of even length \(d\) with an odd number of 1's is \(2^{d-1}\).
   - Applying this lemma, we get that the number of such words in \(B\) is \(2^{n-m}\).

6. **Computing the ratio**:
   - The ratio \(\frac{b_n}{a_n}\) is given by:
     \[
     \frac{b_n}{a_n} = \frac{2^{n-m}}{1} = 2^{n-m}
     \]

The final answer is \(\boxed{2^{n-m}}\).