A box contains 2 red coins, 3 green coins and 4 blue coins. In how many ways can 3 coins be chosen such that at least one coin is green?
Explanation: There are three cases:
- 3 green coins
- 2 green coins + 1 non-green coin
- 1 green coin + 2 non-green coins
Therefore, total number of ways = 3C3 + 3C2 * 6C1 + 3C1 * 6C2 = 1 + 3*6 + 3*15 = 64.
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