# Re: [Scilab-users] Combinatorics

```Hi,

Le 14/04/2022 à 13:12, Claus Futtrup a écrit :
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Dear Scilabers

I hope you can help me out. My combinatorics is a bit rusty.

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So, the spouse has purchased a lock and I wondered how many combinations are available?
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The lock has 10 push buttons, they are numbered 1-2-3-4-5-6-7-8-9-0.

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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.
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I wonder how they are distributed, and how many of the numbers I should activate in the lock to maximize the number of combinations?
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The number of different subsets of k distincts elements of a set composed of n elements is the binomial coefficient (n,k). When n is even, it is maximized for k=n/2. Here
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-->  nchoosek(10, 5)
ans  =

252.

S.

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```Let's see, we have:

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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.
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All (all of the buttons are activated), there's exactly 1 combination for this situation as well (no variability).
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One button pushed. There's obviously 10 possible combinations (push any one of the 10 buttons).
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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?
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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).
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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:
```
1-2-x   8
1-3-x   7
1-4-x   6
1-5-x   5
1-6-x   4
1-7-x   3
1-8-x   2
1-9-0   1

36

```
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:
```
1-x-y   36
2-x-y   28
3-x-y   21
4-x-y   15
5-x-y   10
6-x-y   6
7-x-y   3
8-9-0   1

120

```
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.
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```
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.
```
Best regards,

Claus

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--
Stéphane Mottelet
Ingénieur de recherche
EA 4297 Transformations Intégrées de la Matière Renouvelable
Département Génie des Procédés Industriels
Sorbonne Universités - Université de Technologie de Compiègne
CS 60319, 60203 Compiègne cedex
Tel : +33(0)344234688
http://www.utc.fr/~mottelet
```
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