Among a set of 4 black balls and 3 white balls, how many selections of 3 balls can be made such that at least 2 of them are black balls?
Selecting at least 2 black balls from a set of 4 black balls in total selection of 3 balls can be 2 black and 1 white balls 3 black and 0 white balls Therefore, our solution expression looks 4C2 × 3C1 + 4C3 × 3C0 6 X 3 + 4 X 1 = 18 + 4 = 22 ways
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