There are two white, three black and four red balls.
We have to draw 3 balls, out of these 9 balls in which at least one black ball is included.
So we have following possibilities:
Black balls |
1 |
2 |
3 |
Other than black |
2 |
1 |
0 |
.’. Number of selections = 3C1 x 6C2 + 3C2 x 6C3 + 3C3 x 6C0
= 3×15+ 3×6+1= 45+ 18 + 1= 64