A box contains $2$ white, $3$ black and $4$ red balls.(Balls are of different sizes). In how many ways can $3$ balls be drawn from the box if atleast one black ball is to be included in the draw?

We use the concept of combinations, finding the number of ways to select at least one black ball by computing the difference between total selections and selections with no black balls. The total number of balls is 9, so the total ways to draw 3 balls is 9C3, which equals 84. The number of ways to draw 3 balls from the 6 non-black balls is 6C3, which equals 20. Therefore, the required number of ways is 84 minus 20, giving 64.

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