<html>
<head>
<meta http-equiv="content-type" content="text/html; charset=UTF-8">
</head>
<body>
<p>Dear Scilabers</p>
<p>I hope you can help me out. My combinatorics is a bit rusty.</p>
<p>So, the spouse has purchased a lock and I wondered how many
combinations are available?</p>
<p>The lock has 10 push buttons, they are numbered
1-2-3-4-5-6-7-8-9-0.</p>
<p>From a programming point of view, any of the numbers can be set
on or off, meaning there are 2^10 = 1024 combinations, as far as I
can see.</p>
<p>I wonder how they are distributed, and how many of the numbers I
should activate in the lock to maximize the number of
combinations?</p>
<p>Let's see, we have:</p>
<p>None (none of the buttons are activated), there's exactly 1
combination for this situation. The lock is delivered from the
manufacturer in this state.</p>
<p>All (all of the buttons are activated), there's exactly 1
combination for this situation as well (no variability).</p>
<p>One button pushed. There's obviously 10 possible combinations
(push any one of the 10 buttons).</p>
<p>Two buttons pushed. There's 10 * 9 / 2 = 45 combinations. Each
button can only be pushed once, so once you've selected the first
button, there's only 9 left, but also we divide by two because the
combination are doubled, I mean for example the combination 1-2 =
2-1 ... the lock doesn't know the difference. If you spread out
the possibilities in a 2D plane, it's like ignoring the diagonal
(like pushing the same button twice) and also we either ignore the
upper or lower triangle. Makes sense?</p>
<p>Here starts my trouble. Three buttons pushed. Instead of looking
at a 2D plane, I guess you spread out in 3D. The diagonal line is
more than that - we have several planes where two of the three
numbers are the same (and which are not allowed).</p>
<p>To help myself out, I've tried to write all combinations where
one of the push buttons is number 1. We select all combinations
with the second button being either 2-3-4 and so on, and how many
combinations do we then have for the third option? See table
below:</p>
<p> </p>
<table width="128" cellspacing="0" cellpadding="0" border="0">
<colgroup><col style="width:48pt" width="64" span="2"> </colgroup><tbody>
<tr style="height:15.0pt" height="20">
<td style="height:15.0pt;width:48pt" width="64" height="20">1-2-x</td>
<td style="width:48pt" width="64" align="right">8</td>
</tr>
<tr style="height:15.0pt" height="20">
<td style="height:15.0pt" height="20">1-3-x</td>
<td align="right">7</td>
</tr>
<tr style="height:15.0pt" height="20">
<td style="height:15.0pt" height="20">1-4-x</td>
<td align="right">6</td>
</tr>
<tr style="height:15.0pt" height="20">
<td style="height:15.0pt" height="20">1-5-x</td>
<td align="right">5</td>
</tr>
<tr style="height:15.0pt" height="20">
<td style="height:15.0pt" height="20">1-6-x</td>
<td align="right">4</td>
</tr>
<tr style="height:15.0pt" height="20">
<td style="height:15.0pt" height="20">1-7-x</td>
<td align="right">3</td>
</tr>
<tr style="height:15.0pt" height="20">
<td style="height:15.0pt" height="20">1-8-x</td>
<td align="right">2</td>
</tr>
<tr style="height:15.0pt" height="20">
<td style="height:15.0pt" height="20">1-9-0</td>
<td align="right">1</td>
</tr>
<tr style="height:15.0pt" height="20">
<td style="height:15.0pt" height="20"><br>
</td>
<td align="right">36</td>
</tr>
</tbody>
</table>
<p>We can then do the same for the first button = number 2, and we
get : 7 + 6 + 5 + 4 + 3 + 2 + 1 = 28 combinations and so on. We
get:<br>
</p>
<p> </p>
<table width="128" cellspacing="0" cellpadding="0" border="0">
<colgroup><col style="width:48pt" width="64" span="2"> </colgroup><tbody>
<tr style="height:15.0pt" height="20">
<td style="height:15.0pt;width:48pt" width="64" height="20">1-x-y</td>
<td style="width:48pt" width="64" align="right">36</td>
</tr>
<tr style="height:15.0pt" height="20">
<td style="height:15.0pt" height="20">2-x-y</td>
<td align="right">28</td>
</tr>
<tr style="height:15.0pt" height="20">
<td style="height:15.0pt" height="20">3-x-y</td>
<td align="right">21</td>
</tr>
<tr style="height:15.0pt" height="20">
<td style="height:15.0pt" height="20">4-x-y</td>
<td align="right">15</td>
</tr>
<tr style="height:15.0pt" height="20">
<td style="height:15.0pt" height="20">5-x-y</td>
<td align="right">10</td>
</tr>
<tr style="height:15.0pt" height="20">
<td style="height:15.0pt" height="20">6-x-y</td>
<td align="right">6</td>
</tr>
<tr style="height:15.0pt" height="20">
<td style="height:15.0pt" height="20">7-x-y</td>
<td align="right">3</td>
</tr>
<tr style="height:15.0pt" height="20">
<td style="height:15.0pt" height="20">8-9-0</td>
<td align="right">1</td>
</tr>
<tr style="height:15.0pt" height="20">
<td style="height:15.0pt" height="20"><br>
</td>
<td align="right">120</td>
</tr>
</tbody>
</table>
<p>OK, so that was with three buttons pushed. It's always good to
know the answer (if it's correct :-/ I hope it is), but it's a
tedious process and I was wondering if you could point me to an
easy calculation instead? ... Ideally something that expands to 4
and 5 buttons.</p>
<p>I can 'invent' a calculation, which could be : =10*9*8/(3*2*1) =
120 ... if this indeed shows the internal workings, I'd like to
know why. Sorry my combinatorics is so bad ... I haven't played in
this field for a while.<br>
</p>
Best regards,
<p>Claus<br>
</p>
</body>
</html>