This is an odd question, and I’m not entirely sure that I understand it, but I’ll have a shot.

Let’s split it out into 2 parts,

1, what are the odds that when you remove your 20 cards the flush is in the 20 you picked out are, well the odds that the first card is in there is 20/52, which is the same for each card. So the odds that all of your cards are in the 20 are (20/52)^5

2, Now let’s assume that your cards are in the 20, so you have a chance of picking your flush. We start picking cards randomly. The odds of getting a good card on the first go is 5/20, on the second go it’s 4/19, so we see the odds of getting them all are

5/20 * 4/19 * 3/18 * 2/17 * 1/16

Which is (15! * 5!)/20!

So the odds of doing both step 1 followed by step 2 is

drum roll please

((20/52)^5)*(15! * 5!)/20!)