How does the multiplication principle count costume choices with optional or unlimited accessories?

"Inside a dusty chest marked “Halloween costumes” in the family attic, there are four different outfits (a wizard’s cape, army fatigues, and two others), five different headgears (a batman helmet, a headband, a tiara, etc.), and nine different accessories (a wand, a lightsaber, a pipe, and many others). If a child were to choose a costume by selecting one outfit, one headgear, and one accessory, how many costume choices would he/she have? 4 × 5 × 9 = 180. What if the child were permitted to skip one or more of the items (for instance, choosing a costume with an outfit and accessory, but no headgear)? 5 × 6 × 10 = 300, since now “no choice at all” is effectively another choice for each of the categories. Kind of amazing how much that increases the total! Go back to when the child did have to choose something from each category, but now say they can have any number of accessories (so they could have the wizard’s cape, a batman helmet, plus a lightsaber, pipe, and scepter). Now how many costumes are there? This is 4 × 5 × 2^9, or a whopping 10,240 for those of you keeping score. The 9 changed to a 2^9 because now for each accessory, a costume might include it, or exclude it. That’s two independent choices for each accessory. Note, by the way, that this approach does not work for situations like the license plate example on p. 142. Namely, you can’t say “if a license plate can have fewer than 7 characters, we can just add ‘no character at this position’ as one of the options for that position,” and calculate 37^7 = 94,931,877,133 possible plates. That number is too high. Why? (Hint: for some of those choices, you can get the same license plate in more than one way. Hint 2: if we choose ’A’ for the first license plate character, ‘no character’ for the second, followed by ’NTMAN’ we get the license plate ’ANTMAN’. But what other choices could we have made, that would also have resulted in ’ANTMAN’?)"