You have to pick at least one even factor from the set to make an even product.
There are 3 even numbers to choose from, and we can pick up to 3 additional odd numbers.
For example, if we pick out 1 even number and 2 odd numbers, this can be done in

ways. If we pick out 3 even numbers and 0 odd numbers, this can be done in

way.
The total count is then the sum of all possible selections with at least 1 even number and between 0 and 3 odd numbers.

where we use the binomial identity
