Question

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 atleast one black ball is to be included in the draw?

• A. $64$
• B. 24
• C. $3$
• D. $1$

Answer 1

The correct option is A

$64$

Explanation for the correct answer:

Step 1: Find the number of ways of selecting $3$ balls

We have been given that, a box contains $2$ white balls, $3$ black balls and $4$ red balls.

We need to find the number of ways $3$ balls can be drawn from the box, if atleast one black ball is to be included in the draw.

Since there are $2$ white balls, $3$ black balls and $4$ red balls in a box,

Total balls $=9$

The number of ways of selecting any three balls would be,

$$\begin{gathered} \Rightarrow {}^{9}C_3=\frac{9!}{3!(9-3)!} \\ \Rightarrow {}^{9}C_3=84 \end{gathered}$$

Step 2: Find the number of ways of selecting $3$ balls without black balls.

There would be $6$ balls (if we exclude black balls).

The number of ways of selecting any three balls would be,

$$\begin{gathered} \Rightarrow {}^{6}C_3=\frac{6!}{3!(6-3)!} \\ \Rightarrow {}^{6}C_3=20 \end{gathered}$$

Step 3: Find the number of ways $3$ balls can be drawn from the box if atleast one black ball is to be included in the draw.

Let, there are $\text{'}n\text{'}$ number of ways $3$ balls can be drawn from the box if atleast one black ball is to be included in the draw.

$$\begin{gathered} \Rightarrow n={}^{9}C_3-{}^{6}C_3 \\ \Rightarrow n=84-20 \\ \Rightarrow n=64 \end{gathered}$$

Therefore, option (A) is the correct answer.