<div dir="ltr"><div dir="ltr"><div dir="ltr">Dear Claus,</div><div dir="ltr"><br></div><div dir="ltr">Just in case you have not got an answer yet:</div><div dir="ltr"><br></div><div dir="ltr">> 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 </div><div dir="ltr"><br></div><div>This is right. In general, the formula for the number of k-combinations of n elements is</div><div>n! / ( k! (n-k)! ).</div><div>Alternatively, you can write this as</div><div>n*(n-1)*....*(n-k+1) / (k!).</div><div>(This is the way you have written it above for n = 10 and k = 3.)</div><div>The explanation in general is quite similar to this case:</div><div><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><div>For example, for n=10 and k=4:</div>If you consider "arrangements", i.e., the combinations, where the order matters (e.g., 1-2-5-7 and 2-7-5-1 are different), then you have 10*9*8*7 such arrangements (for the first button you have 10 variants, for the second one 9 variants, etc), but each "combination" (that is, where the order does not matter) is counted 4! (=1*2*3*4) times (number of permutations of 4 elements), so you need to divide: (10*9*8*7)/(4!).</div><div>For more information, see <a href="https://en.wikipedia.org/wiki/Combination" target="_blank">https://en.wikipedia.org/wiki/Combination</a></div><div dir="ltr"><div><div dir="ltr"><div dir="ltr"><div><div dir="ltr"><div><br></div><div>Best regards,</div><div>Mikhail</div><div><br></div></div></div></div></div></div></div></div><br><div class="gmail_quote"><div dir="ltr" class="gmail_attr">Am Do., 14. Apr. 2022 um 13:51 Uhr schrieb Claus Futtrup <<a href="mailto:cfuttrup@gmail.com">cfuttrup@gmail.com</a>>:<br></div><blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex">
<div>
<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:15pt" height="20">
<td style="height:15pt;width:48pt" width="64" height="20">1-2-x</td>
<td style="width:48pt" width="64" align="right">8</td>
</tr>
<tr style="height:15pt" height="20">
<td style="height:15pt" height="20">1-3-x</td>
<td align="right">7</td>
</tr>
<tr style="height:15pt" height="20">
<td style="height:15pt" height="20">1-4-x</td>
<td align="right">6</td>
</tr>
<tr style="height:15pt" height="20">
<td style="height:15pt" height="20">1-5-x</td>
<td align="right">5</td>
</tr>
<tr style="height:15pt" height="20">
<td style="height:15pt" height="20">1-6-x</td>
<td align="right">4</td>
</tr>
<tr style="height:15pt" height="20">
<td style="height:15pt" height="20">1-7-x</td>
<td align="right">3</td>
</tr>
<tr style="height:15pt" height="20">
<td style="height:15pt" height="20">1-8-x</td>
<td align="right">2</td>
</tr>
<tr style="height:15pt" height="20">
<td style="height:15pt" height="20">1-9-0</td>
<td align="right">1</td>
</tr>
<tr style="height:15pt" height="20">
<td style="height:15pt" 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:15pt" height="20">
<td style="height:15pt;width:48pt" width="64" height="20">1-x-y</td>
<td style="width:48pt" width="64" align="right">36</td>
</tr>
<tr style="height:15pt" height="20">
<td style="height:15pt" height="20">2-x-y</td>
<td align="right">28</td>
</tr>
<tr style="height:15pt" height="20">
<td style="height:15pt" height="20">3-x-y</td>
<td align="right">21</td>
</tr>
<tr style="height:15pt" height="20">
<td style="height:15pt" height="20">4-x-y</td>
<td align="right">15</td>
</tr>
<tr style="height:15pt" height="20">
<td style="height:15pt" height="20">5-x-y</td>
<td align="right">10</td>
</tr>
<tr style="height:15pt" height="20">
<td style="height:15pt" height="20">6-x-y</td>
<td align="right">6</td>
</tr>
<tr style="height:15pt" height="20">
<td style="height:15pt" height="20">7-x-y</td>
<td align="right">3</td>
</tr>
<tr style="height:15pt" height="20">
<td style="height:15pt" height="20">8-9-0</td>
<td align="right">1</td>
</tr>
<tr style="height:15pt" height="20">
<td style="height:15pt" 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>
</div>
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</blockquote></div></div>