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?

• A. $64$
• B. $45$
• C. $46$
• D. None of these

Answer 1

The correct option is A

$64$

An explanation for the correct option:

Step- 1: Find the number of possible ways of selecting $3$ balls out of $9$:

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 at least 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.