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A box contains two white, three black and four red balls. In how many ways can three balls be drawn from the box, if atleast one black ball is to be included in the draw.
Solution:
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Verified Answer
Since the box contains 2 white, 3 black and 4 red balls and we have to draw 3 balls out of these 9 balls in which atleast one black ball is included. Hence, we can select the balls in the following ways:
$$
\begin{array}{cccc}
\hline \text { Black balls } & 1 & 2 & 3 \\
\text { Other than black } & 2 & 1 & 0 \\
\hline
\end{array}
$$
$\therefore$ Required number of selections
$$
\begin{aligned}
&={ }^3 C_1 \times{ }^6 C_2+{ }^3 C_2 \times{ }^6 C_1+{ }^3 C_3 \times{ }^6 C_0 \\
&=3 \times 15+3 \times 6+1 \\
&=45+18+1=64
\end{aligned}
$$
$$
\begin{array}{cccc}
\hline \text { Black balls } & 1 & 2 & 3 \\
\text { Other than black } & 2 & 1 & 0 \\
\hline
\end{array}
$$
$\therefore$ Required number of selections
$$
\begin{aligned}
&={ }^3 C_1 \times{ }^6 C_2+{ }^3 C_2 \times{ }^6 C_1+{ }^3 C_3 \times{ }^6 C_0 \\
&=3 \times 15+3 \times 6+1 \\
&=45+18+1=64
\end{aligned}
$$
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