Question

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

• A. $32$
• B. $48$
• C. $64$
• D. $96$

Answer 1

Answer: (C)

$64$

$2$ White balls

$3$ Black balls

$4$ Red balls

Three balls are drawn

Total possible combination

${\Rightarrow }^9{P}_3$

Combination without black

${\Rightarrow }^2{C}_2{\times }^4{C}_1{+}^2{C}_1{\times }^4{C}_2{+}^4{C}_3$

The combination with atleast one black

$\Rightarrow \frac{9\times 8\times 7}{3\times 2\times 1}-[1\times 4+2\times \frac{12}{2}+4]\Rightarrow 12\left(7\right)-\left[20\right]\Rightarrow 84-20=64$

$64$ Combination include atleast one black

Hence the correct answer is $64$