Question

A box contains two white balls, three black balls 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?

Answer 1

Number of white balls = 2

Number of black balls = 3

Number of red balls = 4

Number of balls drawn = 3 balls with at least 1 blackball

We have the following possibilities

White balls (2)	Black  balls (3)	Red balls (4)	Combination
2	1	0	2C2 × 3C1 × 4C0
0	1	2	2C0 × 3C1 × 4C2
1	1	1	2C1 × 3C2 × 4C1
1	2	0	2C1 × 3C2 × 4C0
0	2	1	2C0 × 3C2 × 4C1
0	3	0	2C0 × 3C2 × 4C0

Required number of ways of drawing 3 balls with at least one black ball.

= ²C₂ × ³C₁ × ⁴C₀ +²C₀ × ³C₁ × ⁴C₂ + ²C₁ × ³C₂ × ⁴C₁ + ²C₁ × ³C₂ × ⁴C₀ + ²C₀ × ³C₂ × ⁴C₁ + ²C₀ × ³C₂ × ⁴C₀ 

= `1 xx 3 xx 1 + 1 xx 3 xx (4!)/(2!(4 - 2)!) + 2 xx 3 xx 4 + 2 xx 3 xx 1 + 1 xx 3 xx 4 + 1 xx 1 xx 1`

= `3 + 3 xx (4!)/(2! xx 2!) + 24 + 6 + 12 + 1`

= `3 + 3 xx (4 xx 3 xx 2!)/(2! xx 2!) + 43`

= `3 + 3 xx (4 xx 3)/(2 xx 1) + 43`

= 3 + 18 + 43

= 64