A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one black ball is to be included in the draw?
32
48
64
96
Explanation:
We may have[1 black and 2 non-black] or [2 black and 1 non-black] or [3 black].
| Required number of ways | = (3C1 x 6C2) + (3C2 x 6C1) + (3C3) |
| =[($ 3 \times 5 $ \dfrac{6 \times 5}{2 \times 1} $)]$ + $ [(\dfrac{3 \times 2}{2 \times 1} \times 6]$ + 1 | |
| = (45 + 18 + 1) | |
| = 64. |