How many different lists containing the numbers 1, 4, 5, 8, 17, 21, and nothing else are there in which each odd integer appears before any even integer?

4! * 2!  = 48

Step-by-step explanation:

In general you have 6 elements so there are 6! = 6*5*4*3*2*1  lists in total, now, you have to think about the second condition, an odd integer has to appear before any even integer. Therefore odd integers go first, and since there are 4 odd integers, there are 4! possible lists, and  since there are two even integers there are 2! lists, so in total you have 4! * 2! lists