Permutations & Combinations
Grade 11
Question:
<p>A box contains two white balls, three black balls and four red balls. The number of ways in which three balls can be drawn from the box, so that at least one of the balls is black, is</p>
<p style="display:inline">64</p>
<p style="display:inline">74</p>
<p style="display:inline">84</p>
<p style="display:inline">20</p>
Step-by-Step Solution
Key Concept: Address the 'at least one' requirement by summing the mutually exclusive combinations for picking exactly one, two, and three black balls.
<p>The required number of ways to select</p>
<ol start="1" style="list-style-type: lower-roman;">
<li>1 black ball and 2 others = <sup>3</sup>C<sub>1</sub> <span class="math-tex">\(\cdot\)</span> <sup>6</sup>C<sub>2</sub><br />
= 3 <span class="math-tex">\(\times\)</span> 15 = 45</li>
<li>2 black balls and 1 other = <sup>3</sup>C<sub>2</sub> <span class="math-tex">\(\cdot\)</span> <sup>6</sup>C<sub>1</sub><br />
= 3 <span class="math-tex">\(\times\)</span> 6 = 18</li>
<li>All the three balls are black = <sup>3</sup>C<sub>3</sub> = 1</li>
</ol>
<p>Total number of ways = 45 + 18 + 1 = 64</p>
Correct Answer: A