If we want the product to be positive, there are only 3 possibilities:
5 positive numbers
We choose 5 positive numbers from $ m $ positive numbers: $ \binom{m}{5} $
3 positive numbers and 2 negative numbers
We choose 3 positive numbers from $ m $ positive numbers and 2 negative numbers from $ n $ negative numbers: $ \binom{m}{3} \cdot \binom{n}{2} $
1 positive number and 4 negative numbers
We choose 1 positive number from $ m $ positive numbers and 4 negative numbers from $ n $ negative numbers: $ m \cdot \binom{n}{4} $
Thus, there are $ \binom{m}{5} + \binom{m}{3} \cdot \binom{n}{2} + m \cdot \binom{n}{4} $ such subsets.