Final answer:
The largest possible size of the deck of alphabet cards ensuring that any 1000-card selection contains at least one "A," one "B," and one "C" is 3000 cards. This is determined by using the pigeonhole principle and considering the worst-case scenario for card distribution.
Step-by-step explanation:
To determine the largest possible size of the deck of cards that ensures any collection of 1000 cards contains at least one "A," one "B," and one "C," we can use a combinatorial principle known as the pigeonhole principle. To guarantee that every selection of 1000 includes all three letters, we need to create a situation where removing 999 cards cannot remove all "A"s, "B"s, or "C"s. This means that we require at least 1000-999 = 1 "A," 1 "B," and 1 "C" beyond these 999 cards. Hence, if there were 999 cards without an "A," there must be at least one "A" in the remaining cards. The same goes for "B" and "C."
Realistically, the worst-case scenario is having a 1000-card selection that is just shy of including an "A," a "B," or a "C." We can have up to 999 cards that are not "A," the same for "B," and for "C." So, to guarantee that 1000 cards always include all three letters, there must be 999 non-A's + 999 non-B's + 999 non-C's, plus 1 "A," 1 "B," and 1 "C." Thus, the largest deck would have 999 + 999 + 999 + 3 cards, which totals 3000 cards.