Ok, someone help me?

http://www.pokerstrategy.com/strategy/bss/1608/2/
(When should you fire a 2nd or 3rd barrel? )

page 2, first example, it says:

The ace is an ideal scarecard that we can use for a second barrel in this situation. Let's assume that our opponent will call with AJ and AT and check-raise with 22 or 33. Since he will not always call the flop with AJ or AT, we can further assume that he will do this with exactly half of the possible AJ/AT combinations. For AJo/ATo there are 8 combinations each and 2 for AJs/ATs. Furthermore, 22 is possible and there are 3 combinations for 33.

I'm struggling to understand the number of AJ/AT+ AJs/ATs combos. I think article is saying there are 20 combinations, which fits because when combined with for combinations of 22/33 = 24 combinations which it says villain won't fold. However, I don't understand how we get 20 combinations.

AJo = 12 combinations, AJs = 4 combinations

ATo = 12 combinations, ATs = 4 combinations

and, if villain folds half of these combinations, as it suggests, we get 6+2 for AJ and 6+2 for AT =

**16 total, rather than 20.**
Is it because once we have the board, we discount A

?

but then I think it would be:

AJo = 9 combinations, AJs = 3 combinations

ATo = 9 combinations, ATs = 3 combinations

Giving 24 combinations, half of which would be

**12**
so, I am sure I'm being stupid but if anyone could point out exactly in what way I am being stupid, that would help.

Thanks