6277
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2011-02-01T00:00:01
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<p>The highlight of Gill's fourth birthday party was a game of musical chairs. The game got down to herself, nurse and me. Only two chairs were left - the hard chair and the comfy chair, with a big gap between them. The music stopped and we all piled onto the nearest chair, some on top of one another. If Gill's bottom was firmly in contact with one of the two chairs, in how many different ways
could this have happened?</p>
<p> </p>
<p>If you liked this problem, <a href="http://nrich.maths.org/470">here is an NRICH task</a> which challenges you to use similar mathematical ideas.</p>
<p> </p>
<p> </p>
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<mdoxml version="1.0"><p>Suppose that Gill's bottom was firmly in contact with the Hard Chair. Then the possible combinations would be:<br></br>
</p>
<table style="" border="1">
<tbody>
<tr>
<td> </td>
<td>Hard Chair</td>
<td>Comfy Chair</td>
</tr>
<tr>
<td>1</td>
<td>Gill, Nurse, Me</td>
<td>-</td>
</tr>
<tr>
<td>2</td>
<td>Gill, Me, Nurse</td>
<td>-</td>
</tr>
<tr>
<td>3</td>
<td>Gill, Nurse</td>
<td>Me</td>
</tr>
<tr>
<td>4</td>
<td>Gill, Me</td>
<td>Nurse</td>
</tr>
<tr>
<td>5</td>
<td>Gill</td>
<td>Me, Nurse</td>
</tr>
<tr>
<td>6</td>
<td>Gill</td>
<td>Nurse, Me</td>
</tr>
</tbody>
</table>
<br></br>
<div>So there are $6$ possibilities if Gill sits on the Hard Chair, so there are $2\times 6 = 12$ possibilities in total.</div>
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Weekly Problem 42 - 2008
How many different ways could we have sat on the two remaining musical chairs at Gill's fourth birthday party?
Numbers and the Number System
Counting