A box contains two white balls, three black balls and four red balls. The number of ways in which three balls can be drawn from the box so that at least one of the balls is black is

The box contains a total of 9 balls. The number of ways to draw any 3 balls from the 9 balls is 9C3, which equals (9 times 8 times 7) divided by (3 times 2 times 1), giving 84 ways. The number of ways to draw 3 balls such that none are black means choosing all 3 from the 6 non-black balls, calculated as 6C3, which equals 20 ways. Using the complementary event method, the number of ways to draw at least one black ball is the total ways minus the ways with no black balls: 84 minus 20 equals 64. The required number of ways is 64.

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