Permutations & Combinations
Grade 11
Question:
<p>In how many steps can a 12-step staircase be climbed taking one step or 2 steps at a time?</p>
<p style="display:inline">108</p>
<p style="display:inline">18</p>
<p style="display:inline">232</p>
<p style="display:inline">233</p>
Step-by-Step Solution
Key Concept: Identify all possible combinations of single and double steps that sum to 12 and sum their respective permutations using the formula for arrangements of identical objects.
<table border="1" cellpadding="3" cellspacing="0" style="width:100%;">
<tbody>
<tr>
<td style="text-align: center;"><strong>Number of 2 steps</strong></td>
<td style="text-align: center;"><strong>Number of ways</strong></td>
<td style="text-align: center;"><strong>Reason</strong></td>
</tr>
<tr>
<td style="text-align: center;">0</td>
<td style="text-align: center;">1</td>
<td style="text-align: center;"> </td>
</tr>
<tr>
<td style="text-align: center;">1</td>
<td style="text-align: center;"><span class="math-tex">$\frac{11 !}{10 ! 1 !}$</span> = <sup>11</sup>C<sub>1</sub></td>
<td>There are 10 single steps S and 1 double step D.</td>
</tr>
<tr>
<td style="text-align: center;">2</td>
<td style="text-align: center;"><span class="math-tex">$\frac{10 !}{8 ! 2 !}$</span> = <sup>10</sup>C<sub>2 </sub></td>
<td>There are 8 single steps S and 2 double steps D.</td>
</tr>
<tr>
<td style="text-align: center;">3</td>
<td style="text-align: center;"><span class="math-tex">$\frac{9 !}{3 ! 6 !}$</span> = <sup>9</sup>C<sub>3</sub></td>
<td>There are 6 single steps S and 3 double steps D.</td>
</tr>
<tr>
<td style="text-align: center;">4</td>
<td style="text-align: center;"><span class="math-tex">$\frac{8 !}{4 ! 4 !}$</span> = <sup>8</sup>C<sub>4</sub></td>
<td>There are 4 single steps S and 4 double steps D.</td>
</tr>
<tr>
<td style="text-align: center;">5</td>
<td style="text-align: center;"><span class="math-tex">$\frac{7 !}{5 ! 2 !}$</span> = <sup>7</sup>C<sub>5</sub></td>
<td>There are 2 single steps S and 5 double steps D.</td>
</tr>
<tr>
<td style="text-align: center;">6</td>
<td style="text-align: center;">1</td>
<td> </td>
</tr>
</tbody>
</table>
<p>(10S’s and 1 D are arranged in <span class="math-tex">$\frac{11 !}{10 ! 1 !}$</span> ways.<br />
Applying theorem 3 of permutations)<br />
Total number of ways<br />
= 1 + <sup>11</sup>C<sub>1</sub> + <sup>10</sup>C<sub>2</sub> + <sup>9</sup>C<sub>3</sub> + <sup>8</sup>C<sub>4</sub> + <sup>7</sup>C<sub>5</sub> + 1<br />
= 1 + 11 + 45 + 84 + 70 + 21 + 1 = 233</p>
Correct Answer: D