This is a maximize-minimize question:
We are told that Player C removes only 1 card. Fundamentally, we need to understand what the options are for each player. Remember, each player must remove at least one card. Assuming no rational errors (that is, A mistakenly takes fewer than necessary to win), then the winning situation would be guaranteed if A takes 3 cards at the beginning.
Here's why:
If A gets greedy and removes 5 cards, then B can remove 4 and C takes 1, winning.
If A removes 4 cards, then B can remove 5 cards,
but C will then win. This is what we don't want.
--However, if B removes 4 cards and C takes 1, then A can win.
--If B removes 3 cards and C takes 1, then A can win.
--If B removes 2 cards and C takes 1, then A can win.
--If B removes 1 card and C takes 1, then A can win.
However, if A removes 3 cards or fewer, we get a situation like the following:
--A removes 3 cards, then B removes 5 cards, C removes 1. A can win.
--A removes 3 cards, then B removes 4 cards, C removes 1. A can win.
--A removes 3 cards, then B removes 3 cards, C removes 1. A can win.
--A removes 3 cards, then B removes 2 cards, C removes 1. A can win.
Again, assuming no logical errors on A's part, selecting 4 cards leaves the possibility that C can win, but selecting 3 cards leaves the window open for A.
The answer is C, 3 cards.
_________________
GRE / GMAT Tutor London and Online since 2005. Writer, editor, contributor for test-prep materials.
Struggling with Permutations and Combinations (Combinatorics?). I've put loads of free resources here:
https://privategmattutor.london/gmat-combinatorics-ultimate-guide-to-gmat-permutations-and-combinations/As ever, feel free to get in touch for impartial GRE / GMAT advice or to enquire about lessons at:
https://privategmattutor.london