Question:
A box contains two white, three black and four red balls. In how many ways can three balls be drawn from the box, if at least one black ball is to be included in the draw.
[Hint: Required number of ways =3C1×6C2+3C2x 6C1+3C3.]
Solution:
We know that,
nCr
$=\frac{n !}{r !(n-r) !}$
Drawing 1 black and 2 other ball = 3C1 × 6C2
Drawing 2 black and 1 other ball = 3C2 × 6C1
Drawing 3 black balls = 3C3
Number of ways in which at least one black ball can be drawn =
=(1 black and 2 other )or( 2 black and 1 other )or (3 black)
3C1 × 6C2 + 3C2 × 6C1 + 3C3 =3×15+3×6+1
= 45 + 18 + 1
= 64